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Reverse String characters in java

Am trying to reverse a string using a method in java, I can fetch all the elements of the string and print them out in order via a loop, my problem is reversing the string such that the first comes last and the last comes first, I tried to find a reverse function to no avail... Here is what I have so far...
private static void palindrome() {
char[] name = new char[]{};
String name1;
System.out.println("Enter your name");
Scanner tim = new Scanner(System.in);
name1 = tim.next();
int len = name1.length();
for (int i = 0; i <= len; ++i) {
char b = name1.charAt(i);
System.out.println(b + " ");
}
}
That loop succeeds in printing out the single characters from the string.

You can use StringBuilder like this:
import java.lang.*;
import java.io.*;
import java.util.*;
class ReverseString {
public static void main(String[] args) {
String input = "Geeks for Geeks";
StringBuilder input1 = new StringBuilder();
// append a string into StringBuilder input1
input1.append(input);
// reverse StringBuilder input1
input1 = input1.reverse();
// print reversed String
System.out.println(input1);
}
}
You can also modify your code to do this:
1 -
for (int i = 0; i <= len; ++i) {
char b = name1[len - i];
System.out.println(b + " ");
}
2 -
for (int i = len; i >= 0; --i) {
char b = name1.charAt(i);
System.out.println(b + " ");
}

Using Java 9 codePoints stream you can reverse a string as follows. This example shows the reversal of a string containing surrogate pairs. It works with regular characters as well.
String str = "𝕙𝕖𝕝𝕝𝕠 𝕨𝕠𝕣𝕝𝕕";
String reversed = str.codePoints()
// Stream<String>
.mapToObj(Character::toString)
// concatenate in reverse order
.reduce((a, b) -> b + a)
.get();
System.out.println(reversed); // 𝕕𝕝𝕣𝕠𝕨 𝕠𝕝𝕝𝕖𝕙
See also: Reverse string printing method

You simply need to loop through the array backwards:
for (int i = len - 1; i >= 0; i--) {
char b = name1.charAt(i);
System.out.println(b + " ");
}
You start at the last element which has its index at the position length - 1 and iterate down to the first element (with index zero).
This concept is not specific to Java and also applies to other data structures that provide index based access (such as lists).

Use the built-in reverse() method of the StringBuilder class.
private static void palindrome() {
String name1;
StringBuilder input = new StringBuilder();
System.out.println("Enter your name");
Scanner tim = new Scanner(System.in);
name1 = tim.next();
input.append(name1);
input.reverse();
System.out.println(input);
}

Added reverse() function for your understanding
import java.util.Scanner;
public class P3 {
public static void main(String[] args) {
palindrome();
}
private static void palindrome() {
char[] name = new char[]{};
String name1;
System.out.println("Enter your name");
Scanner tim = new Scanner(System.in);
name1 = tim.next();
String nameReversed = reverse(name1);
int len = name1.length();
for (int i = 0; i < len; ++i) {
char b = name1.charAt(i);
System.out.println(b + " ");
}
}
private static String reverse(String name1) {
char[] arr = name1.toCharArray();
int left = 0, right = arr.length - 1;
while (left < right) {
//swap characters first and last positions
char temp = arr[left];
arr[left++] = arr[right];
arr[right--] = temp;
}
return new String(arr);
}
}

you can try the build-in function charAt()
private String reverseString2(String str) {
if (str == null) {
return null;
}
String result = "";
for (int i = str.length() - 1; i >= 0; i--) {
result = result + str.charAt(i);
}
return result;
}
public void test(){
System.out.println(reverseString2("abcd"));
}
see also rever a string in java

String reversed = new StringBuilder(originalString).reverse().toString();

Common characters in n strings

I m trying to make a function that prints the number of characters common in given n strings. (note that characters may be used multiple times)
I am struggling to perform this operation on n strings However I did it for 2 strings without any characters repeated more than once.
I have posted my code.
public class CommonChars {
public static void main(String[] args) {
String str1 = "abcd";
String str2 = "bcde";
StringBuffer sb = new StringBuffer();
// get unique chars from both the strings
str1 = uniqueChar(str1);
str2 = uniqueChar(str2);
int count = 0;
int str1Len = str1.length();
int str2Len = str2.length();
for (int i = 0; i < str1Len; i++) {
for (int j = 0; j < str2Len; j++) {
// found match stop the loop
if (str1.charAt(i) == str2.charAt(j)) {
count++;
sb.append(str1.charAt(i));
break;
}
}
}
System.out.println("Common Chars Count : " + count + "\nCommon Chars :" +
sb.toString());
}
public static String uniqueChar(String inputString) {
String outputstr="",temp="";
for(int i=0;i<inputstr.length();i++) {
if(temp.indexOf(inputstr.charAt(i))<0) {
temp+=inputstr.charAt(i);
}
}
System.out.println("completed");
return temp;
}
}
3
abcaa
bcbd
bgc
3
their may be chances that a same character can be present multiple times in
a string and you are not supposed to eliminate those characters instead
check the no. of times they are repeated in other strings. for eg
3
abacd
aaxyz
aatre
output should be 2
it will be better if i get solution in java

You have to convert all Strings to Set of Characters and retain all from the first one. Below solution has many places which could be optimised but you should understand general idea.
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
public class Main {
public static void main(String[] args) {
List<String> input = Arrays.asList("jonas", "ton", "bonny");
System.out.println(findCommonCharsFor(input));
}
public static Collection<Character> findCommonCharsFor(List<String> strings) {
if (strings == null || strings.isEmpty()) {
return Collections.emptyList();
}
Set<Character> commonChars = convertStringToSetOfChars(strings.get(0));
strings.stream().skip(1).forEach(s -> commonChars.retainAll(convertStringToSetOfChars(s)));
return commonChars;
}
private static Set<Character> convertStringToSetOfChars(String string) {
if (string == null || string.isEmpty()) {
return Collections.emptySet();
}
Set<Character> set = new HashSet<>(string.length() + 10);
for (char c : string.toCharArray()) {
set.add(c);
}
return set;
}
}
Above code prints:
[n, o]

A better strategy for your problem is to use this method:
public int[] countChars(String s){
int[] count = new int[26];
for(char c: s.toCharArray()){
count[c-'a']++;
}
return count;
}
Now if you have n Strings (String[] strings) just find the min of common chars for each letter:
int[][] result = new int[n][26]
for(int i = 0; i<strings.length;i++){
result[i] = countChars(s);
}
// now if you sum the min common chars for each counter you are ready
int commonChars = 0;
for(int i = 0; i< 26;i++){
int min = result[0][i];
for(int i = 1; i< n;i++){
if(min>result[j][i]){
min = result[j][i];
}
}
commonChars+=min;
}

Get list of characters for each string:
List<Character> chars1 = s1.chars() // list of chars for first string
.mapToObj(c -> (char) c)
.collect(Collectors.toList());
List<Character> chars2 = s2.chars() // list of chars for second string
.mapToObj(c -> (char) c)
.collect(Collectors.toList());
Then use retainAll method:
chars1.retainAll(chars2); // retain in chars1 only the chars that are contained in the chars2 also
System.out.println(chars1.size());
If you want to get number of unique chars just use Collectors.toSet() instead of toList()

Well if one goes for hashing:
public static int uniqueChars(String first, String second) {
boolean[] hash = new boolean[26];
int count = 0;
//reduce first string to unique letters
for (char c : first.toLowerCase().toCharArray()) {
hash[c - 'a'] = true;
}
//reduce to unique letters in both strings
for(char c : second.toLowerCase().toCharArray()){
if(hash[c - 'a']){
count++;
hash[c - 'a'] = false;
}
}
return count;
}
This is using bucketsort which gives a n+m complexity but needs the 26 buckets(the "hash" array).
Imo one can't do better in regards of complexity as you need to look at every letter at least once which sums up to n+m.
Insitu the best you can get is imho somewhere in the range of O(n log(n) ) .
Your aproach is somewhere in the league of O(n²)
Addon: if you need the characters as a String(in essence the same as above with count is the length of the String returned):
public static String uniqueChars(String first, String second) {
boolean[] hash = new boolean[26];
StringBuilder sb = new StringBuilder();
for (char c : first.toLowerCase().toCharArray()) {
hash[c - 'a'] = true;
}
for(char c : second.toLowerCase().toCharArray()){
if(hash[c - 'a']){
sb.append(c);
hash[c - 'a'] = false;
}
}
return sb.toString();
}

public static String getCommonCharacters(String... words) {
if (words == null || words.length == 0)
return "";
Set<Character> unique = words[0].chars().mapToObj(ch -> (char)ch).collect(Collectors.toCollection(TreeSet::new));
for (String word : words)
unique.retainAll(word.chars().mapToObj(ch -> (char)ch).collect(Collectors.toSet()));
return unique.stream().map(String::valueOf).collect(Collectors.joining());
}
Another variant without creating temporary Set and using Character.
public static String getCommonCharacters(String... words) {
if (words == null || words.length == 0)
return "";
int[] arr = new int[26];
boolean[] tmp = new boolean[26];
for (String word : words) {
Arrays.fill(tmp, false);
for (int i = 0; i < word.length(); i++) {
int pos = Character.toLowerCase(word.charAt(i)) - 'a';
if (tmp[pos])
continue;
tmp[pos] = true;
arr[pos]++;
}
}
StringBuilder buf = new StringBuilder(26);
for (int i = 0; i < arr.length; i++)
if (arr[i] == words.length)
buf.append((char)('a' + i));
return buf.toString();
}
Demo
System.out.println(getCommonCharacters("abcd", "bcde")); // bcd

Is reversing a string this way bad practice? [Java] [duplicate]

I have "Hello World" kept in a String variable named hi.
I need to print it, but reversed.
How can I do this? I understand there is some kind of a function already built-in into Java that does that.
Related: Reverse each individual word of “Hello World” string with Java

You can use this:
new StringBuilder(hi).reverse().toString()
StringBuilder was added in Java 5. For versions prior to Java 5, the StringBuffer class can be used instead — it has the same API.

For Online Judges problems that does not allow StringBuilder or StringBuffer, you can do it in place using char[] as following:
public static String reverse(String input){
char[] in = input.toCharArray();
int begin=0;
int end=in.length-1;
char temp;
while(end>begin){
temp = in[begin];
in[begin]=in[end];
in[end] = temp;
end--;
begin++;
}
return new String(in);
}

public static String reverseIt(String source) {
int i, len = source.length();
StringBuilder dest = new StringBuilder(len);
for (i = (len - 1); i >= 0; i--){
dest.append(source.charAt(i));
}
return dest.toString();
}
http://www.java2s.com/Code/Java/Language-Basics/ReverseStringTest.htm

String string="whatever";
String reverse = new StringBuffer(string).reverse().toString();
System.out.println(reverse);

I am doing this by using the following two ways:
Reverse string by CHARACTERS:
public static void main(String[] args) {
// Using traditional approach
String result="";
for(int i=string.length()-1; i>=0; i--) {
result = result + string.charAt(i);
}
System.out.println(result);
// Using StringBuffer class
StringBuffer buffer = new StringBuffer(string);
System.out.println(buffer.reverse());
}
Reverse string by WORDS:
public static void reverseStringByWords(String string) {
StringBuilder stringBuilder = new StringBuilder();
String[] words = string.split(" ");
for (int j = words.length-1; j >= 0; j--) {
stringBuilder.append(words[j]).append(' ');
}
System.out.println("Reverse words: " + stringBuilder);
}

Take a look at the Java 6 API under StringBuffer
String s = "sample";
String result = new StringBuffer(s).reverse().toString();

Here is an example using recursion:
public void reverseString() {
String alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
String reverseAlphabet = reverse(alphabet, alphabet.length()-1);
}
String reverse(String stringToReverse, int index){
if(index == 0){
return stringToReverse.charAt(0) + "";
}
char letter = stringToReverse.charAt(index);
return letter + reverse(stringToReverse, index-1);
}

Here is a low level solution:
import java.util.Scanner;
public class class1 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String inpStr = in.nextLine();
System.out.println("Original String :" + inpStr);
char temp;
char[] arr = inpStr.toCharArray();
int len = arr.length;
for(int i=0; i<(inpStr.length())/2; i++,len--){
temp = arr[i];
arr[i] = arr[len-1];
arr[len-1] = temp;
}
System.out.println("Reverse String :" + String.valueOf(arr));
}
}

I tried, just for fun, by using a Stack. Here my code:
public String reverseString(String s) {
Stack<Character> stack = new Stack<>();
StringBuilder sb = new StringBuilder();
for (int i = 0; i < s.length(); i++) {
stack.push(s.charAt(i));
}
while (!stack.empty()) {
sb.append(stack.pop());
}
return sb.toString();
}

Since the below method (using XOR) to reverse a string is not listed, I am attaching this method to reverse a string.
The Algorithm is based on :
1.(A XOR B) XOR B = A
2.(A XOR B) XOR A = B
Code snippet:
public class ReverseUsingXOR {
public static void main(String[] args) {
String str = "prateek";
reverseUsingXOR(str.toCharArray());
}
/*Example:
* str= prateek;
* str[low]=p;
* str[high]=k;
* str[low]=p^k;
* str[high]=(p^k)^k =p;
* str[low]=(p^k)^p=k;
*
* */
public static void reverseUsingXOR(char[] str) {
int low = 0;
int high = str.length - 1;
while (low < high) {
str[low] = (char) (str[low] ^ str[high]);
str[high] = (char) (str[low] ^ str[high]);
str[low] = (char) (str[low] ^ str[high]);
low++;
high--;
}
//display reversed string
for (int i = 0; i < str.length; i++) {
System.out.print(str[i]);
}
}
}
Output:
keetarp

As others have pointed out the preferred way is to use:
new StringBuilder(hi).reverse().toString()
But if you want to implement this by yourself, I'm afraid that the rest of responses have flaws.
The reason is that String represents a list of Unicode points, encoded in a char[] array according to the variable-length encoding: UTF-16.
This means some code points use a single element of the array (one code unit) but others use two of them, so there might be pairs of characters that must be treated as a single unit (consecutive "high" and "low" surrogates).
public static String reverseString(String s) {
char[] chars = new char[s.length()];
boolean twoCharCodepoint = false;
for (int i = 0; i < s.length(); i++) {
chars[s.length() - 1 - i] = s.charAt(i);
if (twoCharCodepoint) {
swap(chars, s.length() - 1 - i, s.length() - i);
}
twoCharCodepoint = !Character.isBmpCodePoint(s.codePointAt(i));
}
return new String(chars);
}
private static void swap(char[] array, int i, int j) {
char temp = array[i];
array[i] = array[j];
array[j] = temp;
}
public static void main(String[] args) throws Exception {
FileOutputStream fos = new FileOutputStream("C:/temp/reverse-string.txt");
StringBuilder sb = new StringBuilder("Linear B Syllable B008 A: ");
sb.appendCodePoint(65536); //http://unicode-table.com/es/#10000
sb.append(".");
fos.write(sb.toString().getBytes("UTF-16"));
fos.write("\n".getBytes("UTF-16"));
fos.write(reverseString(sb.toString()).getBytes("UTF-16"));
}

Using charAt() method
String name = "gaurav";
String reversedString = "";
for(int i = name.length()-1; i>=0; i--){
reversedString = reversedString + name.charAt(i);
}
System.out.println(reversedString);
Using toCharArray() method
String name = "gaurav";
char [] stringCharArray = name.toCharArray();
String reversedString = "";
for(int i = stringCharArray.length-1; i>=0; i--) {
reversedString = reversedString + stringCharArray[i];
}
System.out.println(reversedString);
Using reverse() method of the Stringbuilder
String name = "gaurav";
String reversedString = new StringBuilder(name).reverse().toString();
System.out.println(reversedString);
Check https://coderolls.com/reverse-a-string-in-java/

It is very simple in minimum code of lines
public class ReverseString {
public static void main(String[] args) {
String s1 = "neelendra";
for(int i=s1.length()-1;i>=0;i--)
{
System.out.print(s1.charAt(i));
}
}
}

This did the trick for me
public static void main(String[] args) {
String text = "abcdefghijklmnopqrstuvwxyz";
for (int i = (text.length() - 1); i >= 0; i--) {
System.out.print(text.charAt(i));
}
}

1. Using Character Array:
public String reverseString(String inputString) {
char[] inputStringArray = inputString.toCharArray();
String reverseString = "";
for (int i = inputStringArray.length - 1; i >= 0; i--) {
reverseString += inputStringArray[i];
}
return reverseString;
}
2. Using StringBuilder:
public String reverseString(String inputString) {
StringBuilder stringBuilder = new StringBuilder(inputString);
stringBuilder = stringBuilder.reverse();
return stringBuilder.toString();
}
OR
return new StringBuilder(inputString).reverse().toString();

System.out.print("Please enter your name: ");
String name = keyboard.nextLine();
String reverse = new StringBuffer(name).reverse().toString();
String rev = reverse.toLowerCase();
System.out.println(rev);
I used this method to turn names backwards and into lower case.

One natural way to reverse a String is to use a StringTokenizer and a stack. Stack is a class that implements an easy-to-use last-in, first-out (LIFO) stack of objects.
String s = "Hello My name is Sufiyan";
Put it in the stack frontwards
Stack<String> myStack = new Stack<>();
StringTokenizer st = new StringTokenizer(s);
while (st.hasMoreTokens()) {
myStack.push(st.nextToken());
}
Print the stack backwards
System.out.print('"' + s + '"' + " backwards by word is:\n\t\"");
while (!myStack.empty()) {
System.out.print(myStack.pop());
System.out.print(' ');
}
System.out.println('"');

public String reverse(String s) {
String reversedString = "";
for(int i=s.length(); i>0; i--) {
reversedString += s.charAt(i-1);
}
return reversedString;
}

public class Test {
public static void main(String args[]) {
StringBuffer buffer = new StringBuffer("Game Plan");
buffer.reverse();
System.out.println(buffer);
}
}

All above solution is too good but here I am making reverse string using recursive programming.
This is helpful for who is looking recursive way of doing reverse string.
public class ReversString {
public static void main(String args[]) {
char s[] = "Dhiral Pandya".toCharArray();
String r = new String(reverse(0, s));
System.out.println(r);
}
public static char[] reverse(int i, char source[]) {
if (source.length / 2 == i) {
return source;
}
char t = source[i];
source[i] = source[source.length - 1 - i];
source[source.length - 1 - i] = t;
i++;
return reverse(i, source);
}
}

You can also try this:
public class StringReverse {
public static void main(String[] args) {
String str = "Dogs hates cats";
StringBuffer sb = new StringBuffer(str);
System.out.println(sb.reverse());
}
}

Procedure :
We can use split() to split the string .Then use reverse loop and add the characters.
Code snippet:
class test
{
public static void main(String args[])
{
String str = "world";
String[] split= str.split("");
String revers = "";
for (int i = split.length-1; i>=0; i--)
{
revers += split[i];
}
System.out.printf("%s", revers);
}
}
//output : dlrow

It gets the value you typed and returns it reversed ;)
public static String reverse (String a){
char[] rarray = a.toCharArray();
String finalvalue = "";
for (int i = 0; i < rarray.length; i++)
{
finalvalue += rarray[rarray.length - 1 - i];
}
return finalvalue;
}

public String reverseWords(String s) {
String reversedWords = "";
if(s.length()<=0) {
return reversedWords;
}else if(s.length() == 1){
if(s == " "){
return "";
}
return s;
}
char arr[] = s.toCharArray();
int j = arr.length-1;
while(j >= 0 ){
if( arr[j] == ' '){
reversedWords+=arr[j];
}else{
String temp="";
while(j>=0 && arr[j] != ' '){
temp+=arr[j];
j--;
}
j++;
temp = reverseWord(temp);
reversedWords+=temp;
}
j--;
}
String[] chk = reversedWords.split(" ");
if(chk == null || chk.length == 0){
return "";
}
return reversedWords;
}
public String reverseWord(String s){
char[] arr = s.toCharArray();
for(int i=0,j=arr.length-1;i<=j;i++,j--){
char tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
return String.valueOf(arr);
}

public static void main(String[] args) {
String str = "Prashant";
int len = str.length();
char[] c = new char[len];
for (int j = len - 1, i = 0; j >= 0; j--, i++) {
c[i] = str.charAt(j);
}
str = String.copyValueOf(c);
System.out.println(str);
}

public void reverString(){
System.out.println("Enter value");
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
try{
String str=br.readLine();
char[] charArray=str.toCharArray();
for(int i=charArray.length-1; i>=0; i--){
System.out.println(charArray[i]);
}
}
catch(IOException ex){
}

recursion:
public String stringReverse(String string) {
if (string == null || string.length() == 0) {
return string;
}
return stringReverse(string.substring(1)) + string.charAt(0);
}

Sequence of characters (or) StringString's Family:
String testString = "Yashwanth#777"; // ~1 1⁄4→D800₁₆«2²⁰
Using Java 8 Stream API
First we convert String into stream by using method CharSequence.chars(), then we use the method IntStream.range to generate a sequential stream of numbers. Then we map this sequence of stream into String.
public static String reverseString_Stream(String str) {
IntStream cahrStream = str.chars();
final int[] array = cahrStream.map( x -> x ).toArray();
int from = 0, upTo = array.length;
IntFunction<String> reverseMapper = (i) -> ( Character.toString((char) array[ (upTo - i) + (from - 1) ]) );
String reverseString = IntStream.range(from, upTo) // for (int i = from; i < upTo ; i++) { ... }
.mapToObj( reverseMapper ) // array[ lastElement ]
.collect(Collectors.joining()) // Joining stream of elements together into a String.
.toString(); // This object (which is already a string!) is itself returned.
System.out.println("Reverse Stream as String : "+ reverseString);
return reverseString;
}
Using a Traditional for Loop
If you want to reverse the string then we need to follow these steps.
Convert String into an Array of Characters.
Iterate over an array in reverse order, append each Character to temporary string variable until the last character.
public static String reverseString( String reverse ) {
if( reverse != null && reverse != "" && reverse.length() > 0 ) {
char[] arr = reverse.toCharArray();
String temp = "";
for( int i = arr.length-1; i >= 0; i-- ) {
temp += arr[i];
}
System.out.println("Reverse String : "+ temp);
}
return null;
}
Easy way to Use reverse method provided form StringBuffer or StringBuilder Classes
StringBuilder and StringBuffer are mutable sequence of characters. That means one can change the value of these object's.
StringBuffer buffer = new StringBuffer(str);
System.out.println("StringBuffer - reverse : "+ buffer.reverse() );
String builderString = (new StringBuilder(str)).reverse().toString;
System.out.println("StringBuilder generated reverse String : "+ builderString );
StringBuffer has the same methods as the StringBuilder, but each method in StringBuffer is synchronized so it is thread safe.

public static String revString(String str){
char[] revCharArr = str.toCharArray();
for (int i=0; i< str.length()/2; i++){
char f = revCharArr[i];
char l = revCharArr[str.length()-i-1];
revCharArr[i] = l;
revCharArr[str.length()-i-1] = f;
}
String revStr = new String(revCharArr);
return revStr;
}

Simple For loop in java
public void reverseString(char[] s) {
int length = s.length;
for (int i = 0; i < s.length / 2; i++) {
// swaping character
char temp = s[length - i - 1];
s[length - i - 1] = s[i];
s[i] = temp;
}
}

How can I split a string into words in Java without using String.split()?

My teacher specifically requested that we split a sentence into words without using String.split(). I've done it using a Vector (which we haven't learned), a while-loop, and substrings. What are other ways of accomplishing this? (preferably without using Vectors/ArrayLists).

I believe that your teacher is asking you to process the string yourself (without using any other libraries to do it for you). Check to see if this is the case - if you can use them, there are things such as StringTokenizer, Pattern, and Scanner to facilitate string processing.
Otherwise...
You will need a list of word separators (such as space, tab, period, etc...) and then walk the array, building a string a character at a time until you hit the word separator. After finding a complete word (you have encountered a word separator character), save it the variable out into your structure (or whatever is required), reset the variable you are building the word in and continue.

Parsing the string character by character, copying each character into a new String, and stopping when you reach a white space character. Then start a new string and continue until you reach the end of the original string.

You can use java.util.StringTokenizer to split a text using desired delimiter. Default delimiter is SPACE/TAB/NEW_LINE.
String myTextToBeSplit = "This is the text to be split into words.";
StringTokenizer tokenizer = new StringTokenizer( myTextToBeSplit );
while ( tokinizer.hasMoreTokens()) {
String word = tokinizer.nextToken();
System.out.println( word ); // word you are looking in
}
As an alternate you can also use java.util.Scanner
Scanner s = new Scanner(myTextToBeSplit).useDelimiter("\\s");
while( s.hasNext() ) {
System.out.println(s.next());
}
s.close();

You can use java.util.Scanner.

import java.util.Arrays;
public class ReverseTheWords {
public static void main(String[] args) {
String s = "hello java how do you do";
System.out.println(Arrays.toString(ReverseTheWords.split(s)));
}
public static String[] split(String s) {
int count = 0;
char[] c = s.toCharArray();
for (int i = 0; i < c.length; i++) {
if (c[i] == ' ') {
count++;
}
}
String temp = "";
int k = 0;
String[] rev = new String[count + 1];
for (int i = 0; i < c.length; i++) {
if (c[i] == ' ') {
rev[k++] = temp;
temp = "";
} else
temp = temp + c[i];
}
rev[k] = temp;
return rev;
}
}

YOu can use StringTokenizer
http://www.java-samples.com/showtutorial.php?tutorialid=236

Or use a Pattern (also known as a regular expression) to try to match the words.

Use a Scanner with ctor (String)
regular expressions and match
StringTokenizer
iterating yourself char by char
recursive iteration

Without using a Vector/List (and without manually re-implementing their ability to re-size themselves for your function), you can take advantage of the simple observation that a string of length N cannot have more than (N+1)/2 words (in integer division). You can declare an array of strings of that size, populate it the same way you populated that Vector, and then copy the results to an array of the size of the number of words you found.
So:
String[] mySplit( String in ){
String[] bigArray = new String[ (in.length()+1)/2 ];
int numWords = 0;
// Populate bigArray with your while loop and keep
// track of the number of words
String[] result = new String[numWords];
// Copy results from bigArray to result
return result;
}

public class MySplit {
public static String[] mySplit(String text,String delemeter){
java.util.List<String> parts = new java.util.ArrayList<String>();
text+=delemeter;
for (int i = text.indexOf(delemeter), j=0; i != -1;) {
parts.add(text.substring(j,i));
j=i+delemeter.length();
i = text.indexOf(delemeter,j);
}
return parts.toArray(new String[0]);
}
public static void main(String[] args) {
String str="012ab567ab0123ab";
String delemeter="ab";
String result[]=mySplit(str,delemeter);
for(String s:result)
System.out.println(s);
}
}

public class sha1 {
public static void main(String[] args) {
String s = "hello java how do you do";
System.out.println(Arrays.toString(sha1.split(s)));
}
public static String[] split(String s) {
int count = 0;
char[] c = s.toCharArray();
for (int i = 0; i < c.length; i++) {
if (c[i] == ' ') {
count++;
}
}
String temp = "";
int k = 0;
String[] rev = new String[count + 1];
for (int i = c.length-1; i >= 0; i--) {
if (c[i] == ' ') {
rev[k++] = temp;
temp = "";
} else
temp = temp + c[i];
}
rev[k] = temp;
return rev;
}
}

Simple touch.! Improve if you want to.
package com.asif.test;
public class SplitWithoutSplitMethod {
public static void main(String[] args) {
split('#',"asif#is#handsome");
}
static void split(char delimeter, String line){
String word = "";
String wordsArr[] = new String[3];
int k = 0;
for(int i = 0; i <line.length(); i++){
if(line.charAt(i) != delimeter){
word+= line.charAt(i);
}else{
wordsArr[k] = word;
word = "";
k++;
}
}
wordsArr[k] = word;
for(int j = 0; j <wordsArr.length; j++)
System.out.println(wordsArr[j]);
}
}

Please try this .
public static String[] mysplit(String mystring) {
String string=mystring+" "; //append " " bcz java string does not hava any ending character
int[] spacetracker=new int[string.length()];// to count no. of spaces in string
char[] array=new char[string.length()]; //store all non space character
String[] tokenArray=new String[string.length()];//to return token of words
int spaceIndex=0;
int parseIndex=0;
int arrayIndex=0;
int k=0;
while(parseIndex<string.length())
{
if(string.charAt(parseIndex)==' '||string.charAt(parseIndex)==' ')
{
spacetracker[spaceIndex]=parseIndex;
spaceIndex++;
parseIndex++;
}else
{
array[arrayIndex]=string.charAt(parseIndex);
arrayIndex++;
parseIndex++;
}
}
for(int i=0;i<spacetracker.length;i++)
{
String token="";
for(int j=k;j<(spacetracker[i])-i;j++)
{
token=token+array[j];
k++;
}
tokenArray[i]=token;
//System.out.println(token);
token="";
}
return tokenArray;
}
Hope this helps

import java.util.*;
class StringSplit {
public static void main(String[] args)
{
String s="splitting a string without using split()";
ArrayList<Integer> al=new ArrayList<Integer>(); //Instead you can also use a String
ArrayList<String> splitResult=new ArrayList<String>();
for(int i=0;i<s.length();i++)
if(s.charAt(i)==' ')
al.add(i);
al.add(0, 0);
al.add(al.size(),s.length());
String[] words=new String[al.size()];
for(int j=0;j<=words.length-2;j++)
splitResult.add(s.substring(al.get(j),al.get(j+1)).trim());
System.out.println(splitResult);
}
}
Time complexity: O(n)

You can use java Pattern to do it in easy way.
package com.company;
import java.util.regex.Pattern;
public class umeshtest {
public static void main(String a[]) {
String ss = "I'm Testing and testing the new feature";
Pattern.compile(" ").splitAsStream(ss).forEach(s -> System.out.println(s));
}
}

You can also use String.substring or charAt[].

Removing duplicates from a String in Java

I am trying to iterate through a string in order to remove the duplicates characters.
For example the String aabbccdef should become abcdef
and the String abcdabcd should become abcd
Here is what I have so far:
public class test {
public static void main(String[] args) {
String input = new String("abbc");
String output = new String();
for (int i = 0; i < input.length(); i++) {
for (int j = 0; j < output.length(); j++) {
if (input.charAt(i) != output.charAt(j)) {
output = output + input.charAt(i);
}
}
}
System.out.println(output);
}
}
What is the best way to do this?

Convert the string to an array of char, and store it in a LinkedHashSet. That will preserve your ordering, and remove duplicates. Something like:
String string = "aabbccdefatafaz";
char[] chars = string.toCharArray();
Set<Character> charSet = new LinkedHashSet<Character>();
for (char c : chars) {
charSet.add(c);
}
StringBuilder sb = new StringBuilder();
for (Character character : charSet) {
sb.append(character);
}
System.out.println(sb.toString());

Using Stream makes it easy.
noDuplicates = Arrays.asList(myString.split(""))
.stream()
.distinct()
.collect(Collectors.joining());
Here is some more documentation about Stream and all you can do with
it :
https://docs.oracle.com/javase/8/docs/api/java/util/stream/package-summary.html
The 'description' part is very instructive about the benefits of Streams.

Try this simple solution:
public String removeDuplicates(String input){
String result = "";
for (int i = 0; i < input.length(); i++) {
if(!result.contains(String.valueOf(input.charAt(i)))) {
result += String.valueOf(input.charAt(i));
}
}
return result;
}

I would use the help of LinkedHashSet. Removes dups (as we are using a Set, maintains the order as we are using linked list impl). This is kind of a dirty solution. there might be even a better way.
String s="aabbccdef";
Set<Character> set=new LinkedHashSet<Character>();
for(char c:s.toCharArray())
{
set.add(Character.valueOf(c));
}

Create a StringWriter. Run through the original string using charAt(i) in a for loop. Maintain a variable of char type keeping the last charAt value. If you iterate and the charAt value equals what is stored in that variable, don't add to the StringWriter. Finally, use the StringWriter.toString() method and get a string, and do what you need with it.

Here is an improvement to the answer by Dave.
It uses HashSet instead of the slightly more costly LinkedHashSet, and reuses the chars buffer for the result, eliminating the need for a StringBuilder.
String string = "aabbccdefatafaz";
char[] chars = string.toCharArray();
Set<Character> present = new HashSet<>();
int len = 0;
for (char c : chars)
if (present.add(c))
chars[len++] = c;
System.out.println(new String(chars, 0, len)); // abcdeftz

Java 8 has a new String.chars() method which returns a stream of characters in the String. You can use stream operations to filter out the duplicate characters like so:
String out = in.chars()
.mapToObj(c -> Character.valueOf((char) c)) // bit messy as chars() returns an IntStream, not a CharStream (which doesn't exist)
.distinct()
.map(Object::toString)
.collect(Collectors.joining(""));

String input = "AAAB";
String output = "";
for (int index = 0; index < input.length(); index++) {
if (input.charAt(index % input.length()) != input
.charAt((index + 1) % input.length())) {
output += input.charAt(index);
}
}
System.out.println(output);
but you cant use it if the input has the same elements, or if its empty!

Code to remove the duplicate characters in a string without using any additional buffer. NOTE: One or two additional variables are fine. An extra array is not:
import java.util.*;
public class Main{
public static char[] removeDupes(char[] arr){
if (arr == null || arr.length < 2)
return arr;
int len = arr.length;
int tail = 1;
for(int x = 1; x < len; x++){
int y;
for(y = 0; y < tail; y++){
if (arr[x] == arr[y]) break;
}
if (y == tail){
arr[tail] = arr[x];
tail++;
}
}
return Arrays.copyOfRange(arr, 0, tail);
}
public static char[] bigArr(int len){
char[] arr = new char[len];
Random r = new Random();
String alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890!##$%^&*()-=_+[]{}|;:',.<>/?`~";
for(int x = 0; x < len; x++){
arr[x] = alphabet.charAt(r.nextInt(alphabet.length()));
}
return arr;
}
public static void main(String args[]){
String result = new String(removeDupes(new char[]{'a', 'b', 'c', 'd', 'a'}));
assert "abcd".equals(result) : "abcda should return abcd but it returns: " + result;
result = new String(removeDupes(new char[]{'a', 'a', 'a', 'a'}));
assert "a".equals(result) : "aaaa should return a but it returns: " + result;
result = new String(removeDupes(new char[]{'a', 'b', 'c', 'a'}));
assert "abc".equals(result) : "abca should return abc but it returns: " + result;
result = new String(removeDupes(new char[]{'a', 'a', 'b', 'b'}));
assert "ab".equals(result) : "aabb should return ab but it returns: " + result;
result = new String(removeDupes(new char[]{'a'}));
assert "a".equals(result) : "a should return a but it returns: " + result;
result = new String(removeDupes(new char[]{'a', 'b', 'b', 'a'}));
assert "ab".equals(result) : "abba should return ab but it returns: " + result;
char[] arr = bigArr(5000000);
long startTime = System.nanoTime();
System.out.println("2: " + new String(removeDupes(arr)));
long endTime = System.nanoTime();
long duration = (endTime - startTime);
System.out.println("Program took: " + duration + " nanoseconds");
System.out.println("Program took: " + duration/1000000000 + " seconds");
}
}
How to read and talk about the above code:
The method called removeDupes takes an array of primitive char called arr.
arr is returned as an array of primitive characters "by value". The arr passed in is garbage collected at the end of Main's member method removeDupes.
The runtime complexity of this algorithm is O(n) or more specifically O(n+(small constant)) the constant being the unique characters in the entire array of primitive chars.
The copyOfRange does not increase runtime complexity significantly since it only copies a small constant number of items. The char array called arr is not stepped all the way through.
If you pass null into removeDupes, the method returns null.
If you pass an empty array of primitive chars or an array containing one value, that unmodified array is returned.
Method removeDupes goes about as fast as physically possible, fully utilizing the L1 and L2 cache, so Branch redirects are kept to a minimum.
A 2015 standard issue unburdened computer should be able to complete this method with an primitive char array containing 500 million characters between 15 and 25 seconds.
Explain how this code works:
The first part of the array passed in is used as the repository for the unique characters that are ultimately returned. At the beginning of the function the answer is: "the characters between 0 and 1" as between 0 and tail.
We define the variable y outside of the loop because we want to find the first location where the array index that we are looking at has been duplicated in our repository. When a duplicate is found, it breaks out and quits, the y==tail returns false and the repository is not contributed to.
when the index x that we are peeking at is not represented in our repository, then we pull that one and add it to the end of our repository at index tail and increment tail.
At the end, we return the array between the points 0 and tail, which should be smaller or equal to in length to the original array.
Talking points exercise for coder interviews:
Will the program behave differently if you change the y++ to ++y? Why or why not.
Does the array copy at the end represent another 'N' pass through the entire array making runtime complexity O(n*n) instead of O(n) ? Why or why not.
Can you replace the double equals comparing primitive characters with a .equals? Why or why not?
Can this method be changed in order to do the replacements "by reference" instead of as it is now, "by value"? Why or why not?
Can you increase the efficiency of this algorithm by sorting the repository of unique values at the beginning of 'arr'? Under which circumstances would it be more efficient?

public class RemoveRepeated4rmString {
public static void main(String[] args) {
String s = "harikrishna";
String s2 = "";
for (int i = 0; i < s.length(); i++) {
Boolean found = false;
for (int j = 0; j < s2.length(); j++) {
if (s.charAt(i) == s2.charAt(j)) {
found = true;
break; //don't need to iterate further
}
}
if (found == false) {
s2 = s2.concat(String.valueOf(s.charAt(i)));
}
}
System.out.println(s2);
}
}

public static void main(String a[]){
String name="Madan";
System.out.println(name);
StringBuilder sb=new StringBuilder(name);
for(int i=0;i<name.length();i++){
for(int j=i+1;j<name.length();j++){
if(name.charAt(i)==name.charAt(j)){
sb.deleteCharAt(j);
}
}
}
System.out.println("After deletion :"+sb+"");
}

import java.util.Scanner;
public class dublicate {
public static void main(String... a) {
System.out.print("Enter the String");
Scanner Sc = new Scanner(System.in);
String st=Sc.nextLine();
StringBuilder sb=new StringBuilder();
boolean [] bc=new boolean[256];
for(int i=0;i<st.length();i++)
{
int index=st.charAt(i);
if(bc[index]==false)
{
sb.append(st.charAt(i));
bc[index]=true;
}
}
System.out.print(sb.toString());
}
}

To me it looks like everyone is trying way too hard to accomplish this task. All we are concerned about is that it copies 1 copy of each letter if it repeats. Then because we are only concerned if those characters repeat one after the other the nested loops become arbitrary as you can just simply compare position n to position n + 1. Then because this only copies things down when they're different, to solve for the last character you can either append white space to the end of the original string, or just get it to copy the last character of the string to your result.
String removeDuplicate(String s){
String result = "";
for (int i = 0; i < s.length(); i++){
if (i + 1 < s.length() && s.charAt(i) != s.charAt(i+1)){
result = result + s.charAt(i);
}
if (i + 1 == s.length()){
result = result + s.charAt(i);
}
}
return result;
}

String str1[] ="Hi helloo helloo oooo this".split(" ");
Set<String> charSet = new LinkedHashSet<String>();
for (String c: str1)
{
charSet.add(c);
}
StringBuilder sb = new StringBuilder();
for (String character : charSet)
{
sb.append(character);
}
System.out.println(sb.toString());

I think working this way would be more easy,,,
Just pass a string to this function and the job is done :) .
private static void removeduplicate(String name)
{ char[] arr = name.toCharArray();
StringBuffer modified =new StringBuffer();
for(char a:arr)
{
if(!modified.contains(Character.toString(a)))
{
modified=modified.append(Character.toString(a)) ;
}
}
System.out.println(modified);
}

public class RemoveDuplicatesFromStingsMethod1UsingLoops {
public static void main(String[] args) {
String input = new String("aaabbbcccddd");
String output = "";
for (int i = 0; i < input.length(); i++) {
if (!output.contains(String.valueOf(input.charAt(i)))) {
output += String.valueOf(input.charAt(i));
}
}
System.out.println(output);
}
}
output: abcd

You can't. You can create a new String that has duplicates removed. Why aren't you using StringBuilder (or StringBuffer, presumably)?
You can run through the string and store the unique characters in a char[] array, keeping track of how many unique characters you've seen. Then you can create a new String using the String(char[], int, int) constructor.
Also, the problem is a little ambiguous—does “duplicates” mean adjacent repetitions? (In other words, what should happen with abcab?)

Oldschool way (as we wrote such a tasks in Apple ][ Basic, adapted to Java):
int i,j;
StringBuffer str=new StringBuffer();
Scanner in = new Scanner(System.in);
System.out.print("Enter string: ");
str.append(in.nextLine());
for (i=0;i<str.length()-1;i++){
for (j=i+1;j<str.length();j++){
if (str.charAt(i)==str.charAt(j))
str.deleteCharAt(j);
}
}
System.out.println("Removed non-unique symbols: " + str);

Here is another logic I'd like to share. You start comparing from midway of the string length and go backward.
Test with:
input = "azxxzy";
output = "ay";
String removeMidway(String input){
cnt = cnt+1;
StringBuilder str = new StringBuilder(input);
int midlen = str.length()/2;
for(int i=midlen-1;i>0;i--){
for(int j=midlen;j<str.length()-1;j++){
if(str.charAt(i)==str.charAt(j)){
str.delete(i, j+1);
midlen = str.length()/2;
System.out.println("i="+i+",j="+j+ ",len="+ str.length() + ",midlen=" + midlen+ ", after deleted = " + str);
}
}
}
return str.toString();
}

Another possible solution, in case a string is an ASCII string, is to maintain an array of 256 boolean elements to denote ASCII character appearance in a string. If a character appeared for the first time, we keep it and append to the result. Otherwise just skip it.
public String removeDuplicates(String input) {
boolean[] chars = new boolean[256];
StringBuilder resultStringBuilder = new StringBuilder();
for (Character c : input.toCharArray()) {
if (!chars[c]) {
resultStringBuilder.append(c);
chars[c] = true;
}
}
return resultStringBuilder.toString();
}
This approach will also work with Unicode string. You just need to increase chars size.

Solution using JDK7:
public static String removeDuplicateChars(final String str){
if (str == null || str.isEmpty()){
return str;
}
final char[] chArray = str.toCharArray();
final Set<Character> set = new LinkedHashSet<>();
for (char c : chArray) {
set.add(c);
}
final StringBuilder sb = new StringBuilder();
for (Character character : set) {
sb.append(character);
}
return sb.toString();
}

String str = "eamparuthik#gmail.com";
char[] c = str.toCharArray();
String op = "";
for(int i=0; i<=c.length-1; i++){
if(!op.contains(c[i] + ""))
op = op + c[i];
}
System.out.println(op);

public static String removeDuplicateChar(String str){
char charArray[] = str.toCharArray();
StringBuilder stringBuilder= new StringBuilder();
for(int i=0;i<charArray.length;i++){
int index = stringBuilder.toString().indexOf(charArray[i]);
if(index <= -1){
stringBuilder.append(charArray[i]);
}
}
return stringBuilder.toString();
}

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class RemoveDuplicacy
{
public static void main(String args[])throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.print("Enter any word : ");
String s = br.readLine();
int l = s.length();
char ch;
String ans=" ";
for(int i=0; i<l; i++)
{
ch = s.charAt(i);
if(ch!=' ')
ans = ans + ch;
s = s.replace(ch,' '); //Replacing all occurrence of the current character by a space
}
System.out.println("Word after removing duplicate characters : " + ans);
}
}

public static void main(String[] args) {
int i,j;
StringBuffer str=new StringBuffer();
Scanner in = new Scanner(System.in);
System.out.print("Enter string: ");
str.append(in.nextLine());
for (i=0;i<str.length()-1;i++)
{
for (j=1;j<str.length();j++)
{
if (str.charAt(i)==str.charAt(j))
str.deleteCharAt(j);
}
}
System.out.println("Removed String: " + str);
}

This is improvement on solution suggested by #Dave. Here, I am implementing in single loop only.
Let's reuse the return of set.add(T item) method and add it simultaneously in StringBuffer if add is successfull
This is just O(n). No need to make a loop again.
String string = "aabbccdefatafaz";
char[] chars = string.toCharArray();
StringBuilder sb = new StringBuilder();
Set<Character> charSet = new LinkedHashSet<Character>();
for (char c : chars) {
if(charSet.add(c) ){
sb.append(c);
}
}
System.out.println(sb.toString()); // abcdeftz

Simple solution is to iterate through the given string and put each unique character into another string(in this case, a variable result ) if this string doesn't contain that particular character.Finally return result string as output.
Below is working and tested code snippet for removing duplicate characters from the given string which has O(n) time complexity .
private static String removeDuplicate(String s) {
String result="";
for (int i=0 ;i<s.length();i++) {
char ch = s.charAt(i);
if (!result.contains(""+ch)) {
result+=""+ch;
}
}
return result;
}
If the input is madam then output will be mad.
If the input is anagram then output will be angrm
Hope this helps.
Thanks

For the simplicity of the code- I have taken hardcore input, one can take input by using Scanner class also
public class KillDuplicateCharInString {
public static void main(String args[]) {
String str= "aaaabccdde ";
char arr[]= str.toCharArray();
int n = arr.length;
String finalStr="";
for(int i=0;i<n;i++) {
if(i==n-1){
finalStr+=arr[i];
break;
}
if(arr[i]==arr[i+1]) {
continue;
}
else {
finalStr+=arr[i];
}
}
System.out.println(finalStr);
}
}

public static void main (String[] args)
{
Scanner sc = new Scanner(System.in);
String s = sc.next();
String str = "";
char c;
for(int i = 0; i < s.length(); i++)
{
c = s.charAt(i);
str = str + c;
s = s.replace(c, ' ');
if(i == s.length() - 1)
{
System.out.println(str.replaceAll("\\s", ""));
}
}
}

package com.st.removeduplicate;
public class RemoveDuplicate {
public static void main(String[] args) {
String str1="shushil",str2="";
for(int i=0; i<=str1.length()-1;i++) {
int count=0;
for(int j=0;j<=i;j++) {
if(str1.charAt(i)==str1.charAt(j))
count++;
if(count >1)
break;
}
if(count==1)
str2=str2+str1.charAt(i);
}
System.out.println(str2);
}
}