How can I check to see if a String contains a whitespace character, an empty space or " ". If possible, please provide a Java example.
For example: String = "test word";
For checking if a string contains whitespace use a Matcher and call its find method.
Pattern pattern = Pattern.compile("\\s");
Matcher matcher = pattern.matcher(s);
boolean found = matcher.find();
If you want to check if it only consists of whitespace then you can use String.matches:
boolean isWhitespace = s.matches("^\\s*$");
Check whether a String contains at least one white space character:
public static boolean containsWhiteSpace(final String testCode){
if(testCode != null){
for(int i = 0; i < testCode.length(); i++){
if(Character.isWhitespace(testCode.charAt(i))){
return true;
}
}
}
return false;
}
Reference:
Character.isWhitespace(char)
Using the Guava library, it's much simpler:
return CharMatcher.WHITESPACE.matchesAnyOf(testCode);
CharMatcher.WHITESPACE is also a lot more thorough when it comes to Unicode support.
This will tell if you there is any whitespaces:
Either by looping:
for (char c : s.toCharArray()) {
if (Character.isWhitespace(c)) {
return true;
}
}
or
s.matches(".*\\s+.*")
And StringUtils.isBlank(s) will tell you if there are only whitepsaces.
Use Apache Commons StringUtils:
StringUtils.containsWhitespace(str)
public static void main(String[] args) {
System.out.println("test word".contains(" "));
}
You could use Regex to determine if there's a whitespace character. \s.
More info of regex here.
Use this code, was better solution for me.
public static boolean containsWhiteSpace(String line){
boolean space= false;
if(line != null){
for(int i = 0; i < line.length(); i++){
if(line.charAt(i) == ' '){
space= true;
}
}
}
return space;
}
You can use charAt() function to find out spaces in string.
public class Test {
public static void main(String args[]) {
String fav="Hi Testing 12 3";
int counter=0;
for( int i=0; i<fav.length(); i++ ) {
if(fav.charAt(i) == ' ' ) {
counter++;
}
}
System.out.println("Number of spaces "+ counter);
//This will print Number of spaces 4
}
}
String str = "Test Word";
if(str.indexOf(' ') != -1){
return true;
} else{
return false;
}
Maybe I'm late with the most updated answer. You can use one of the following solution:
public static boolean containsWhiteSpace(final String input) {
if (isNotEmpty(input)) {
for (int i = 0; i < input.length(); i++) {
if (Character.isWhitespace(input.charAt(i)) || Character.isSpaceChar(input.charAt(i))) {
return true;
}
}
}
return false;
}
or
public static boolean containsWhiteSpace(final String input) {
return CharMatcher.whitespace().matchesAnyOf(input);
}
import java.util.Scanner;
public class camelCase {
public static void main(String[] args)
{
Scanner user_input=new Scanner(System.in);
String Line1;
Line1 = user_input.nextLine();
int j=1;
//Now Read each word from the Line and convert it to Camel Case
String result = "", result1 = "";
for (int i = 0; i < Line1.length(); i++) {
String next = Line1.substring(i, i + 1);
System.out.println(next + " i Value:" + i + " j Value:" + j);
if (i == 0 | j == 1 )
{
result += next.toUpperCase();
} else {
result += next.toLowerCase();
}
if (Character.isWhitespace(Line1.charAt(i)) == true)
{
j=1;
}
else
{
j=0;
}
}
System.out.println(result);
Use org.apache.commons.lang.StringUtils.
to search for whitespaces
boolean withWhiteSpace = StringUtils.contains("my name", " ");
To delete all whitespaces in a string
StringUtils.deleteWhitespace(null) = null
StringUtils.deleteWhitespace("") = ""
StringUtils.deleteWhitespace("abc") = "abc"
StringUtils.deleteWhitespace(" ab c ") = "abc"
I purpose to you a very simple method who use String.contains:
public static boolean containWhitespace(String value) {
return value.contains(" ");
}
A little usage example:
public static void main(String[] args) {
System.out.println(containWhitespace("i love potatoes"));
System.out.println(containWhitespace("butihatewhitespaces"));
}
Output:
true
false
package com.test;
public class Test {
public static void main(String[] args) {
String str = "TestCode ";
if (str.indexOf(" ") > -1) {
System.out.println("Yes");
} else {
System.out.println("Noo");
}
}
}
Related
I'm making a Palindrome Generator. Basically the user inputs a word or sentence and the program outputs whether or not its a Palindrome, which is a word that is spelled the same forwards and backwards like "wow" or "racecar". My program works fine, however the output text will repeat itself like fifty times and I can't seem to figure out where the issue is without messing everything up. Help would be appreciated.
import javax.swing.JOptionPane;
public class palindromedectector {
public static void main(String[] args) {
String testStrings = "";
testStrings = JOptionPane.showInputDialog("Enter word: ");
for (int i = 0; i < testStrings.length(); i++)
{
System.out.print("\"" + testStrings + "\"");
if (isPalindrome(stripString(testStrings)))
System.out.println(" is a palindrome.");
else
System.out.println(" is not a palindrome.");
}
}
public static String stripString(String strip)
{
strip = strip.toUpperCase();
String stripped= "";
for (int i= 0; i< strip.length(); i++)
{
if (Character.isLetter(strip.charAt(i)))
stripped += strip.charAt(i);
}
return stripped;
}
public static boolean isPalindrome (String str)
{
boolean status = false;
if (str.length() <= 1)
status = true;
else if (str.charAt(0) == str.charAt(str.length()-1))
{
status = isPalindrome (str.substring(1, str.length()-1));
}
return status;
}
}
Main issue is that you run isPalindrome check for the same string in the loop, probably you wanted to run multiple checks
public static void main(String[] args) {
final int attempts = 5;
for (int i = 0; i < attempts; i++) {
String word = JOptionPane.showInputDialog("Enter word: ");
System.out.print("\"" + word + "\"");
if (isPalindrome(stripString(word))) {
System.out.println(" is a palindrome.");
} else {
System.out.println(" is not a palindrome.");
}
}
}
Also, the main functionality may be implemented in a shorter way:
// use regexp to get rid of non-letters
private static String stripString(String word) {
if (null == word || word.isEmpty()) {
return word;
}
return word.replaceAll("[^A-Za-z]", "").toUpperCase(); // remove all non-letters
}
// use Java Stream API to check letters using half of word length
private static boolean isPalindrome(String word) {
if (null == word) {
return false;
}
final int len = word.length();
if (len < 2) {
return true;
}
return IntStream.range(0, len/2)
.allMatch(i -> word.charAt(i) == word.charAt(len - 1 - i));
}
Basic problem: You are testing if the word is a palindrome testStrings.length() times, ie once for every letter in the word, rather than just once.
Remove the for loop in your main() method.
public static void main(String[] args)
{
Scanner sentence = new Scanner(System.in);
System.out.println("This program will determine if an inputted phrase is a palindrome.");
System.out.println(" ");
System.out.println("Enter a phrase, word, or sentence:");
String a = sentence.nextLine();
String b = a.toLowerCase().replaceAll("[^a-z]"," "); //as long as the words are spelt the same way, the caps don't matter and it ignores spaces and punctuation
System.out.println();
System.out.println(palindromeChecker(b)); //calls method
}
public static String palindromeChecker(String b)
{
String reverse = new StringBuilder(b).reverse().toString();
String c;
if(b.equals(reverse)) {
c = "The word " +b+ " is a palindrome"; }
else {
c = "The word " +b+ " is not a palindrome"; }
return c;
}
}
My, problem is that for example, if i do Eva, can I see bees in a cave? It should be a palindrome, however it's not can u please help me with this and please try not to make it complicated.
Replace:
String b = a.toLowerCase().replaceAll("[^a-z]"," ");
With
String b = a.toLowerCase().replaceAll("[^a-z]","");
Otherwise, you're replacing non-alphabetical characters with spaces, which can influence the checking of the reverse String.
You need to remove all non-letters from your string:
public class Palindrome {
public static String strip(String s) {
return s.toLowerCase().replaceAll("[^a-z]", "");
}
public static boolean isPalindrome(String s) {
for (int i = 0; i < s.length() / 2; i++) {
if (s.charAt(i) != s.charAt(s.length() - 1 - i)) {
return false;
}
}
return true;
}
public static void main(String[] args) {
System.out.println(strip("Eva, can I see bees in a cave?"));
System.out.println(isPalindrome(strip("Eva, can I see bees in a cave?")));
}
}
Output:
evacaniseebeesinacave
true
public class PalindromeStringWithReverse {
public static void main(String[] args) {
String str = "21raceca r12";
str = str.replaceAll(" ", "");
boolean isPalindrome = false;
for (int i = 0; i < str.length() / 2; i++) {
if (str.charAt(i) == str.charAt((str.length() - 1) - i)) {
isPalindrome = true;
continue;
}
isPalindrome = false;
}
if (isPalindrome) {
System.out.println("Palindrome");
} else {
System.out.println("not palindrome");
}
}
}
I need to replace a repeated char with $% followed by the char followed by $%.
e.g. "HELLO" will become "HE$%L$%O"
The following code that I wrote gives "HE$%L$%LO".
Please guide
int index=0;
String str1="";
String str2="";
String str4="";
String str5="";
for(int i=0;i<str.length();i++) {
char ch=str.charAt(i);
index=str.indexOf(ch);
if(index!=i) {
str4="$%"+str.charAt(index)+ "$%";
str1=str.charAt(index)+str5;
str2=str.replaceFirst(str1,str4);
}
}
return str2;
It looks like there's code missing because i can't see the duplicate character check, but what you want to do is go through str5 before you concat it and strip off all of the duplicate characters that are at the beginning. Then concat to your String.
Here a solution: Id solves the case if duplicates are more than 2 too. So remove all duplicates:
public class Converter {
public static void main(String[] args) {
final String result = replace("HELLO");
System.out.println("result = " + result);
}
private static String replace(String data) {
final StringBuilder stringBuilder = new StringBuilder();
for (int i = 0; i < data.length();) {
int j = i + 1;
while (j < data.length() && data.charAt(i) == data.charAt(j)) {
j++;
}
if(j > i + 1) { // exist duplicate
stringBuilder.append("$%").append(data.charAt(i)).append("$%");
} else {
stringBuilder.append(data.charAt(i));
}
i = j;
}
return stringBuilder.toString();
}
}
And the result is:
result = HE$%L$%O
I need to write a static method that takes a String as a parameter and returns a new String obtained by replacing every instance of repeated adjacent letters with a single instance of that letter without using regular expressions. For example if I enter "maaaakkee" as a String, it returns "make".
I already tried the following code, but it doesn't seem to display the last character.
Here's my code:
import java.util.Scanner;
public class undouble {
public static void main(String [] args){
Scanner console = new Scanner(System.in);
System.out.println("enter String: ");
String str = console.nextLine();
System.out.println(removeSpaces(str));
}
public static String removeSpaces(String str){
String ourString="";
int j = 0;
for (int i=0; i<str.length()-1 ; i++){
j = i+1;
if(str.charAt(i)!=str.charAt(j)){
ourString+=str.charAt(i);
}
}
return ourString;
}
}
You could use regular expressions for that.
For instance:
String input = "ddooooonnneeeeee";
System.out.println(input.replaceAll("(.)\\1{1,}", "$1"));
Output:
done
Pattern explanation:
"(.)\\1{1,}" means any character (added to group 1) followed by itself at least once
"$1" references contents of group 1
maybe:
for (int i=1; i<str.length() ; i++){
j = i+1;
if(str.charAt(i)!=str.charAt(j)){
ourString+=str.charAt(i);
}
}
The problem is with your condition. You say compare i and i+1 in each iteration and in last iteration you have both i and j pointing to same location so it will never print the last character. Try this unleass you want to use regex to achive this:
EDIT:
public void removeSpaces(String str){
String ourString="";
for (int i=0; i<str.length()-1 ; i++){
if(i==0){
ourString = ""+str.charAt(i);
}else{
if(str.charAt(i-1) != str.charAt(i)){
ourString = ourString +str.charAt(i);
}
}
}
System.out.println(ourString);
}
if you cannot use replace or replaceAll, here is an alternative. O(2n), O(N) for stockage and O(N) for creating the string. It removes all repeated chars in the string put them in a stringbuilder.
input : abcdef , output : abcdef
input : aabbcdeef, output : cdf
private static String remove_repeated_char(String str)
{
StringBuilder result = new StringBuilder();
HashMap<Character, Integer> items = new HashMap<>();
for (int i = 0; i < str.length(); i++)
{
Character current = str.charAt(i);
Integer ocurrence = items.get(current);
if (ocurrence == null)
items.put(current, 1);
else
items.put(current, ocurrence + 1);
}
for (int i = 0; i < str.length(); i++)
{
Character current = str.charAt(i);
Integer ocurrence = items.get(current);
if (ocurrence == 1)
result.append(current);
}
return result.toString();
}
import java.util.*;
public class string2 {
public static void main(String[] args) {
//removes repeat character from array
Scanner sc=new Scanner(System.in);
StringBuffer sf=new StringBuffer();
System.out.println("enter a string");
sf.append(sc.nextLine());
System.out.println("string="+sf);
int i=0;
while( i<sf.length())
{
int j=1+i;
while(j<sf.length())
{
if(sf.charAt(i)==sf.charAt(j))
{
sf.deleteCharAt(j);
}
else
{
j=j+1;
}
}
i=i+1;
}
System.out.println("string="+sf);
}
}
Input AABBBccDDD, Output BD
Input ABBCDDA, Outout C
private String reducedString(String s){
char[] arr = s.toCharArray();
String newString = "";
Map<Character,Integer> map = new HashMap<Character,Integer>();
map.put(arr[0],1);
for(int index=1;index<s.length();index++)
{
Character key = arr[index];
int value;
if(map.get(key) ==null)
{
value =0;
}
else
{
value = map.get(key);
}
value = value+1;
map.put(key,value);
}
Set<Character> keyset = map.keySet();
for(Character c: keyset)
{
int value = map.get(c);
if(value%2 !=0)
{
newString+=c;
}
}
newString = newString.equals("")?"Empty String":newString;
return newString;
}
public class RemoveDuplicateCharecterInString {
static String input = new String("abbbbbbbbbbbbbbbbccccd");
static String output = "";
public static void main(String[] args)
{
// TODO Auto-generated method stub
for (int i = 0; i < input.length(); i++) {
char temp = input.charAt(i);
boolean check = false;
for (int j = 0; j < output.length(); j++) {
if (output.charAt(j) == input.charAt(i)) {
check = true;
}
}
if (!check) {
output = output + input.charAt(i);
}
}
System.out.println(" " + output);
}
}
Answer : abcd
public class RepeatedChar {
public static void main(String[] args) {
String rS = "maaaakkee";
String outCome= rS.charAt(0)+"";
int count =0;
char [] cA =rS.toCharArray();
for(int i =0; i+1<cA.length; ++i) {
if(rS.charAt(i) != rS.charAt(i+1)) {
outCome += rS.charAt(i+1);
}
}
System.out.println(outCome);
}
}
TO WRITE JAVA PROGRAM TO REMOVE REPEATED CHARACTERS:
package replace;
public class removingrepeatedcharacters
{
public static void main(String...args){
int i,j=0,count=0;
String str="noordeen";
String str2="noordeen";
char[] ch=str.toCharArray();
for(i=0;i<=5;i++)
{
count=0;
for(j=0;j<str2.length();j++)
{
if(ch[i]==str2.charAt(j))
{
count++;
System.out.println("at the index "+j +"position "+ch[i]+ "+ count is"+count);
if(count>=2){
str=str2;
str2=str.replaceFirst(Character.toString(ch[j]),Character.toString(' '));
}
System.out.println("after replacing " +str2);
}
}
}
}
}
String outstr = "";
String outstring = "";
for(int i = 0; i < str.length() - 1; i++) {
if(str.charAt(i) != str.charAt(i + 1)) {
outstr = outstr + str.charAt(i);
}
outstring = outstr + str.charAt(i);
}
System.out.println(outstring);
public static void remove_duplicates(String str){
String outstr="";
String outstring="";
for(int i=0;i<str.length()-1;i++) {
if(str.charAt(i)!=str.charAt(i+1)) {
outstr=outstr+str.charAt(i);
}
outstring=outstr+str.charAt(i);
}
System.out.println(outstring);
}
More fun with java 7:
System.out.println("11223344445555".replaceAll("(?<nums>.+)\\k<nums>+","${nums}"));
No more cryptic numbers in regexes.
public static String removeDuplicates(String str) {
String str2 = "" + str.charAt(0);
for (int i = 1; i < str.length(); i++) {
if (str.charAt(i - 1) == str.charAt(i) && i != 0) {
continue;
}
str2 = str2 + str.charAt(i);
}
return str2;
}
This question already has answers here:
Check string for palindrome
(42 answers)
Closed 7 years ago.
I want to check if a string is a palindrome or not. I would like to learn an easy method to check the same using least possible string manipulations
Using reverse is overkill because you don't need to generate an extra string, you just need to query the existing one. The following example checks the first and last characters are the same, and then walks further inside the string checking the results each time. It returns as soon as s is not a palindrome.
The problem with the reverse approach is that it does all the work up front. It performs an expensive action on a string, then checks character by character until the strings are not equal and only then returns false if it is not a palindrome. If you are just comparing small strings all the time then this is fine, but if you want to defend yourself against bigger input then you should consider this algorithm.
boolean isPalindrome(String s) {
int n = s.length();
for (int i = 0; i < (n/2); ++i) {
if (s.charAt(i) != s.charAt(n - i - 1)) {
return false;
}
}
return true;
}
For the least lines of code and the simplest case
if(s.equals(new StringBuilder(s).reverse().toString())) // is a palindrome.
Here is a simple one"
public class Palindrome {
public static void main(String [] args){
Palindrome pn = new Palindrome();
if(pn.isPalindrome("ABBA")){
System.out.println("Palindrome");
} else {
System.out.println("Not Palindrome");
}
}
public boolean isPalindrome(String original){
int i = original.length()-1;
int j=0;
while(i > j) {
if(original.charAt(i) != original.charAt(j)) {
return false;
}
i--;
j++;
}
return true;
}
}
You can try something like this :
String variable = ""; #write a string name
StringBuffer rev = new StringBuffer(variable).reverse();
String strRev = rev.toString();
if(variable.equalsIgnoreCase(strRev)) # Check the condition
Here's a good class :
public class Palindrome {
public static boolean isPalindrome(String stringToTest) {
String workingCopy = removeJunk(stringToTest);
String reversedCopy = reverse(workingCopy);
return reversedCopy.equalsIgnoreCase(workingCopy);
}
protected static String removeJunk(String string) {
int i, len = string.length();
StringBuffer dest = new StringBuffer(len);
char c;
for (i = (len - 1); i >= 0; i--) {
c = string.charAt(i);
if (Character.isLetterOrDigit(c)) {
dest.append(c);
}
}
return dest.toString();
}
protected static String reverse(String string) {
StringBuffer sb = new StringBuffer(string);
return sb.reverse().toString();
}
public static void main(String[] args) {
String string = "Madam, I'm Adam.";
System.out.println();
System.out.println("Testing whether the following "
+ "string is a palindrome:");
System.out.println(" " + string);
System.out.println();
if (isPalindrome(string)) {
System.out.println("It IS a palindrome!");
} else {
System.out.println("It is NOT a palindrome!");
}
System.out.println();
}
}
Enjoy.
public boolean isPalindrom(String text) {
StringBuffer stringBuffer = new StringBuffer(text);
return stringBuffer.reverse().toString().equals(text);
}
I guess this is simple way to check palindrome
String strToRevrse = "MOM";
strToRevrse.equalsIgnoreCase(new StringBuilder(strToRevrse).reverse().toString());
I'm new to java and I'm taking up your question as a challenge to improve my knowledge as well so please forgive me if this does not answer your question well:
import java.util.ArrayList;
import java.util.List;
public class PalindromeRecursiveBoolean {
public static boolean isPalindrome(String str) {
str = str.toUpperCase();
char[] strChars = str.toCharArray();
List<Character> word = new ArrayList<>();
for (char c : strChars) {
word.add(c);
}
while (true) {
if ((word.size() == 1) || (word.size() == 0)) {
return true;
}
if (word.get(0) == word.get(word.size() - 1)) {
word.remove(0);
word.remove(word.size() - 1);
} else {
return false;
}
}
}
}
If the string is made of no letters or just one letter, it is a
palindrome.
Otherwise, compare the first and last letters of the string.
If the first and last letters differ, then the string is not a palindrome
Otherwise, the first and last letters are the same. Strip them from the string, and determine whether the string that remains is a palindrome. Take the answer for this smaller string and use it as the answer for the original string then repeat from 1.
The only string manipulation is changing the string to uppercase so that you can enter something like 'XScsX'
check this condition
String string="//some string...//"
check this...
if(string.equals((string.reverse())
{
it is palindrome
}
public static boolean istPalindrom(char[] word){
int i1 = 0;
int i2 = word.length - 1;
while (i2 > i1) {
if (word[i1] != word[i2]) {
return false;
}
++i1;
--i2;
}
return true;
}
import java.util.Scanner;
public class FindAllPalindromes {
static String longestPalindrome;
public String oldPalindrome="";
static int longest;
public void allSubstrings(String s){
for(int i=0;i<s.length();i++){
for(int j=1;j<=s.length()-i;j++){
String subString=s.substring(i, i+j);
palindrome(subString);
}
}
}
public void palindrome(String sub){
System.out.println("String to b checked is "+sub);
StringBuilder sb=new StringBuilder();
sb.append(sub); // append string to string builder
sb.reverse();
if(sub.equals(sb.toString())){ // palindrome condition
System.out.println("the given String :"+sub+" is a palindrome");
longestPalindrome(sub);
}
else{
System.out.println("the string "+sub+"iss not a palindrome");
}
}
public void longestPalindrome(String s){
if(s.length()>longest){
longest=s.length();
longestPalindrome=s;
}
else if (s.length()==longest){
oldPalindrome=longestPalindrome;
longestPalindrome=s;
}
}
public static void main(String[] args) {
FindAllPalindromes fp=new FindAllPalindromes();
Scanner sc=new Scanner(System.in);
System.out.println("Enter the String ::");
String s=sc.nextLine();
fp.allSubstrings(s);
sc.close();
if(fp.oldPalindrome.length()>0){
System.out.println(longestPalindrome+"and"+fp.oldPalindrome+":is the longest palindrome");
}
else{
System.out.println(longestPalindrome+":is the longest palindrome`````");
}}
}