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  1. Prepare
  2. Java
  3. Strings
  4. Java String Reverse
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Java String Reverse

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1939 Discussions

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  • sawkhedkaraditi
     + 0 comments

    import java.io.; import java.util.;

    public class Solution {

    public static void main(String[] args) {

    Scanner sc=new Scanner(System.in);
    String A=sc.next();
    /* Enter your code here. Print output to STDOUT. */
    StringBuffer sb = new StringBuffer(A);
    sb.reverse();
    if(A.equals(sb.toString()))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }

   sc.close(); 
}

    }

  • 19551a04e2
     + 0 comments

    public static void main(String[] args) {

        Scanner sc=new Scanner(System.in);
    String A=sc.next();
    int len = 0;
    /* Enter your code here. Print output to STDOUT. */
    for(int i = A.length()-1; i>=(A.length())/2; i--){
        if(A.charAt(i) != A.charAt(len)){
            System.out.println("No");
            return;
        }
        len++;
    }
    System.out.println("Yes");
}

  • serenitydiver
     + 0 comments

    Java 8 Based 

      // Not optimized version
	// Time Complexity: O(n)
	 // Space Complexity: O(n)  due to creating a substring
	 private static boolean isPalindrome(String input) {
        int inputLength = input.length();
        String secondHalfOfInput = input.substring(inputLength / 2);
        int secondHalfOfInputLength = secondHalfOfInput.length();
        for(int i = 0; i < inputLength / 2; i++) {
            if(input.charAt(i) != secondHalfOfInput.charAt(secondHalfOfInputLength - (i + 1))) {
                return false;
            }
        }

        return true;
    }
    
  //  optimized version
	// Time Complexity: O(n)
	 // Space Complexity: O(1)  No substring		
    private static boolean isPalindromeOptimised(String input) {
        int left = 0;
        int right = input.length() - 1;
        while (left < right) {
            if (input.charAt(left) != input.charAt(right)) {
                return false;
            }
            left++;
            right--;
        }
        return true;
  }

  • abramowicz
     + 0 comments

    The previous one-liner solution is awsome! My idea was to use "converging iterators" approach that works in-place without copying the data (sub-strings, concatenation, etc.). What's more, this approach is very fast as it may give an answer even before reading the whole string, possibly after comparing just two letters!

    import java.util.Scanner;

public class Solution {
    private static boolean isPalindrome(String str) {
        for (int f = 0, r = str.length()-1; f < r; ++f, --r) {
            if (str.charAt(f) != str.charAt(r))
                return false;
        }
        return true;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str = sc.nextLine();
        sc.close();
        System.out.println(isPalindrome(str) ? "Yes" : "No");
    }
}

  • md_pasha9370
     + 0 comments

    Java Logic just one line

        String check =  new StringBuffer(A).reverse().toString();