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Java, extract integer values within defined characters from long string

I have a program where I receive a long string in the format
characters$xxx,characters$xx,characters$xx, (....)
x is some digit of some integer with an arbitrary number of digits. The integer values are always contained within $ and ,.
I need to extract the integers into an integer array then print that array. The second part is easy, but how to extract those integers?
an example string: adsdfsh$1234,khjdfd$356,hsgadfsd$98,ghsdsk$4623,
the arraay should contain 1234, 356, 98, 4623
below is my basic logic
import java.util.Scanner;
class RandomStuff {
public static void main (String[]args){
Scanner keyboard = new Scanner(System.in);
String input = keyboard.next();
int count =0;
// counts number of $, because $ will always preceed an int in my string
for(int i=0;i<input.length();i++ ){
if (input.charAt(i)=='$')
count++;}
/* also I'm traversing the string twice here so my complexity is at least
o(2n) if someone knows how to reduce that, please tell me*/
int [] intlist = new int[count];
// fill the array
int arrayindex =0;
for (int i=0; i<input.length();i++){
if (input.charAt(i)=='$'){
/*insert all following characters as a single integer in intlist[arrayindex]
until we hit the character ','*/}
if (input.charAt(i)==','){
arrayindex++;
/*stop recording characters*/}
}
// i can print an array so I'll just omit the rest
keyboard.close();
}

You can use a regular expression with a positive lookbehind to find all consecutive sequences of digits preceded by a $ symbol. Matcher#results can be used to get all of the matches.
String str = "adsdfsh$1234,khjdfd$356,hsgadfsd$98,ghsdsk$4623";
int[] nums = Pattern.compile("(?<=\\$)\\d+").matcher(str).results()
.map(MatchResult::group)
.mapToInt(Integer::parseInt).toArray();
System.out.println(Arrays.toString(nums));

It can done like this
var digitStarts = new ArrayList<Integer>()
var digitsEnds = new ArrayList<Integer>()
// Get the start and end of each digit
for (int i=0; i<input.length();i++){
if(input[i] == '$' ) digitsStarts.add(i)
if(input[i] == ',') digitEnds.add(i)
}
// Get the digits as strings
var digitStrings = new ArrayList<String>()
for(int i=0;i<digitsStart.length; i++ ) {
digitsString.add(input.substring(digitsStarts[i]+1,digitEnds[i]))
}
// Convert to Int
var digits = new ArrayList<Int>
for(int i=0;i<digitString;i++) {
digits.add(Integer.valueOf(digitStrings[i]))
}

In a very simple way:
public static void main(String[] args) {
String str = "adsdfsh$1234,khjdfd$356,hsgadfsd$98,ghsdsk$4623";
String strArray[] = str.split(",");
int numbers[] = new int[strArray.length];
int j = 0;
for(String s : strArray) {
numbers[j++] = Integer.parseInt(s.substring(s.indexOf('$')+1));
}
for(j=0;j<numbers.length;j++)
System.out.print(numbers[j]+" ");
}
OUTPUT: 1234 356 98 4623

Unique characters from string

I am trying to print all the unique characters from a string but I am not getting proper output. Also, I want to check if someone enters integer in string, I want to print Invalid String. How can I achieve this?
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
char[] ch = new char[20];
System.out.println("Enter the sentence:");
String sent = sc.nextLine().replaceAll(" ", "");
int count = 0;
for (int i = 0; i < sent.length(); i++) {
int j = (sent.length() - 1);
count = 0;
while (j > i) {
if (sent.charAt(j) == sent.charAt(i)) {
sent = deleteCharAt(sent, i);
sent = deleteCharAt(sent, j - 1);
break;
}
j--;
}
}
for (int i = 0; i < sent.length(); i++) {
System.out.println(sent.charAt(i));
}
}
private static String deleteCharAt(String strValue, int index) {
return strValue.substring(0, index) + strValue.substring(index + 1);
}
Enter the sentence:
java is good object oriented programming language
a
v
i
s
o
d
b
c
r
e
e
d
p
g
m
m
n
l
u

You are probably going to want to use a Set. These data structures are like Lists, except:
They do not have an ordering (meaning you cant call set.get(3))
They do not allow duplicates
You can think of them as a Map without any values.
If you have a String and you want to get all the unique chars from it. The steps are as follows:
String string = "hello"; // 4 unique characters
Set<Character> uniqueChars = new HashSet<>(); // create an empty set to put the unique chars into
// split into char[]
char[] chars = string.toCharArray();
Arrays.stream(chars).forEach(c -> {
// the following code will be run once for every char in the array
uniqueChars.add(c);
// adding the same char twice does not insert it twice
});
This can be written more concisely as:
String string = "hello";
Set<Character> uniqueChars = new HashSet<>();
Arrays.stream(string.toCharArray()).forEach(uniqueChars::add); // using a Java 8 method reference
If you want to reject any char that is a numerical digit, you can use the following line:
boolean containsDigit = Arrays.stream(string.toCharArray())
.filter(Character::isDigit) // filter out all the non digit characters
.findAny() // check if there are any remaining
.isPresent();

It would be more efficient and easier to read code to use a HashSet:
HashSet<Character> h = new HashSet<Character>();
for (int i = 0; i <= (sent.length() - 1); i++)
h.add(sent.charAt(i));
Iterator<Character> i = h.iterator();
while (i.hasNext())
System.out.println(i.next());

To avoid duplicate you can use a Set: is a collection that doesn't allow duplicates. You have to use a specific implementation of that interface, such as HashSet. You can do something like this:
public class UniqueChar {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("Enter the sentence:");
String withoutSpaces = sc.nextLine();
withoutSpaces = "asc34csf"; // mock example
Set<Character> goodChars = new HashSet<>();
String sent = withoutSpaces.replaceAll(" ", "");
int count = 0;
for (int i = 0; i < sent.length(); i++) {
char currChar = sent.charAt(i);
// do not add a character if is a digit
if(Character.isDigit(currChar))
System.out.println("Digit!");
else
goodChars.add(currChar); // add a character only if not present
}
String output = "";
for (Character character : goodChars) {
output += character; // concat in a single output string
}
System.out.println(output);
}
So you simply jump characters that are digits, if it's a character it'll be added to the collection (and the Set manages internally the fact that if it's a duplicate, it will not be added), then concat the elements of the Set in a single String.
You can find more information about HashSet in the Java documentation.

You can write your code something like this:
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int[] alpha = new int[26];
System.out.println("Enter the sentence:");
String sent = sc.nextLine().replaceAll(" ", "");
for (int i = 0; i < sent.length(); i++) {
int temp = sent[i] - 'a';
if (temp >= 0 && temp <= 25) {
alpha[temp] = 1;
} else {
System.out.println("Invalid String");
return;
}
}
for (int i = 0; i < 26; i++) {
if (alpha[i] == 1)
System.out.println((char) (i + 'a'));
}
}
This uses an array of length 26 as the workaround if you don't want to use any SET like data structure.
This code can bring you problem if your sentence have capital letters as well. You can avoid that problem by using toLowercase like function just before the loop.

Generate Random String in java [duplicate]

This question already has answers here:
How to generate a random alpha-numeric string
(46 answers)
Closed 6 years ago.
I'm trying to generate a string between capital A-Z in java using Secure Random. Currently I'm able to generate an alphanumeric string with special characters but I want a string with only upper case alphabets.
public String createRandomCode(int codeLength, String id){
char[] chars = id.toCharArray();
StringBuilder sb = new StringBuilder();
Random random = new SecureRandom();
for (int i = 0; i < codeLength; i++) {
char c = chars[random.nextInt(chars.length)];
sb.append(c);
}
String output = sb.toString();
System.out.println(output);
return output ;
}
The input parameters are length of the output string & id whhich is alphanumeric string.Can't understand what modifications to make to the above code to generate only upper case alphabet string. Please help..

Your method randomly selects characters out of the id argument. If you want those to only be uppercase letters, then pass a string with those characters:
String randomCode = createRandomCode(length, "ABCDEFGHIJKLMNOPQRSTUVWXYZ");
EDIT If you want to avoid duplicates, you can't just select characters at random. You'll want to shuffle them and pick out the first n characters:
public String createRandomCode(int codeLength, String id) {
List<Character> temp = id.chars()
.mapToObj(i -> (char)i)
.collect(Collectors.toList());
Collections.shuffle(temp, new SecureRandom());
return temp.stream()
.map(Object::toString)
.limit(codeLength)
.collect(Collectors.joining());
}
EDIT 2 Just for fun, here's another way to implement the original random code generator (allowing duplicates):
public static String createRandomCode(int codeLength, String id) {
return new SecureRandom()
.ints(codeLength, 0, id.length())
.mapToObj(id::charAt)
.map(Object::toString)
.collect(Collectors.joining());
}

Here is generator that I wrote and use:
public class RandomGenerator {
private static final String characters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
public static String generateRandom(int length) {
Random random = new SecureRandom();
if (length <= 0) {
throw new IllegalArgumentException("String length must be a positive integer");
}
StringBuilder sb = new StringBuilder(length);
for (int i = 0; i < length; i++) {
sb.append(characters.charAt(random.nextInt(characters.length())));
}
return sb.toString();
}
}
in numChars string you can put any characters you want to be included. int length parameter is the length of generated random string.

Here is an example method that uses the int range for characters A to Z (also this method avoids duplicate characters in the String) :
public String createRandomCode(final int codeLength) {
int min = 65;// A
int max = 90;// Z
StringBuilder sb = new StringBuilder();
Random random = new SecureRandom();
for (int i = 0; i < codeLength; i++) {
Character c;
do {
c = (char) (random.nextInt((max - min) + 1) + min);
} while (sb.indexOf(c.toString()) > -1);
sb.append(c);
}
String output = sb.toString();
System.out.println(output);
return output;
}
The range part comes from this topic : Generating random integers in a specific range

java replace correct number into letter

Well, this is my first time get here.
I'm trying to figure out the correct way to replace number into letter.
In this case, I need two steps.
First, convert letter to number. Second, restore number to word.
Words list: a = 1, b = 2, f = 6 and k = 11.
I have word: "b a f k"
So, for first step, it must be: "2 1 6 11"
Number "2 1 6 11" must be converted to "b a f k".
But, I failed at second step.
Code I've tried:
public class str_number {
public static void main(String[] args){
String word = "b a f k";
String number = word.replace("a", "1").replace("b","2").replace("f","6").replace("k","11");
System.out.println(word);
System.out.println(number);
System.out.println();
String text = number.replace("1", "a").replace("2","b").replace("6","f").replace("11","k");
System.out.println(number);
System.out.println(text);
}
}
Result:
b a f k
2 1 6 11
2 1 6 11
b a f aa
11 must be a word "k", but it's converted to "aa"
What is the right way to fix this?
Or do you have any other ways to convert letter to number and vice versa?
Thank you.

It would be good to write methods for conversion between number and letter format. I would write some code like this and use it generally instead of hard coding replace each time.
public class test {
static ArrayList <String> letter = new ArrayList<String> ();
static ArrayList <String> digit = new ArrayList<String> ();
public static void main(String[] args) {
createTable();
String test="b a f k";
String test1="2 1 6 11";
System.out.println(letterToDigit(test));
System.out.println(digitToLetter(test1));
}
public static void createTable()
{
//Create all your Letter to number Mapping here.
//Add all the letters and digits
letter.add("a");
digit.add("1");
letter.add("b");
digit.add("2");
letter.add("c");
digit.add("3");
letter.add("d");
digit.add("4");
letter.add("e");
digit.add("5");
letter.add("f");
digit.add("6");
letter.add("g");
digit.add("7");
letter.add("h");
digit.add("8");
letter.add("i");
digit.add("9");
letter.add("j");
digit.add("10");
letter.add("k");
digit.add("11");
letter.add("l");
digit.add("12");
letter.add("m");
digit.add("13");
letter.add("n");
digit.add("14");
letter.add("o");
digit.add("14");
letter.add("p");
digit.add("15");
//Carry so on till Z
}
public static String letterToDigit(String input)
{
String[] individual = input.split(" ");
String result="";
for(int i=0;i<individual.length;i++){
if(letter.contains(individual[i])){
result+=Integer.toString(letter.indexOf(individual[i])+1)+ " ";
}
}
return result;
}
public static String digitToLetter(String input)
{
String[] individual = input.split(" ");
String result="";
for(int i=0;i<individual.length;i++){
if(digit.contains(individual[i])){
result+=letter.get(digit.indexOf(individual[i])) + " ";
}
}
return result;
}
}

I would actually not use replace in this case.
A more generic solution would be to simply convert it to a char and subtract the char a from it.
int n = word.charAt(0) - 'a' + 1;
This should return an int with the value you are looking for.
If you want this to be an string you can easily do
String s = Integer.parseInt(word.charAt(0) - 'a' + 1);
And as in your case you are doing a whole string looping through the length of it and changing all would give you the result
String s = "";
for(int i = 0; i < word.length(); i++) {
if(s.charAt(i) != ' ') {
s = s + Integer.toString(word.charAt(i) - 'a' + 1) + " ";
}
}
and then if you want this back to an String with letters instead
String text = "";
int temp = 0;
for(int i = 0; i < s.length(); i++) {
if(s.charAt(i) == ' ') {
text = text + String.valueOf((char) (temp + 'a' - 1));
temp = 0;
} else if {
temp = (temp*10)+Character.getNumericValue(s.charAt(i));
}
}

You can just reverse the replacement:
String text = number.replace("11","k").replace("2","b").replace("6","f").replace("1","a");
Simplest solution IMO.
When adding other numbers, first replace these with two digits, then these with one.

Replace this:
String text = number.replace("1", "a").replace("2","b").replace("6","f").replace("11","k");
By this:
String text = number.replace("11","k").replace("1", "a").replace("2","b").replace("6","f");
Right now, the first replace you're doing: ("1", "a")
is invalidating the last one: ("11","k")

I think you would need to store the number as an array of ints. Otherwise, there is no way of knowing if 11 is aa or k. I would create a Map and then loop over the characters in the String. You could have one map for char-to-int and one for int-to-char.
Map<Character,Integer> charToIntMap = new HashMap<Character,Integer>();
charToIntMap.put('a',1);
charToIntMap.put('b',2);
charToIntMap.put('f',6);
charToIntMap.put('k',11);
Map<Integer,Character> intToCharMap = new HashMap<Integer,Character>();
intToCharMap.put(1,'a');
intToCharMap.put(2,'b');
intToCharMap.put(6,'f');
intToCharMap.put(11,'k');
String testStr = "abfk";
int[] nbrs = new int[testStr.length()];
for(int i = 0; i< testStr.length(); i++ ){
nbrs[i] = charToIntMap.get(testStr.charAt(i));
}
StringBuilder sb = new StringBuilder();
for(int num : nbrs){
sb.append(num);
}
System.out.println(sb.toString());
//Reverse
sb = new StringBuilder();
for(int i=0; i<nbrs.length; i++){
sb.append(intToCharMap.get(nbrs[i]));
}
System.out.println(sb.toString());

This failed because the replace("1", "a") replaced both 1s with a characters. The quickest fix is to perform the replace of all the double-digit numbers first, so there are no more double-digit numbers left when the single-digit numbers get replaced.
String text = number.replace("11","k").replace("1", "a").
replace("2","b").replace("6","f");

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);
}
}