Single line solution (ok, maybe 2 lines):

System.out.println( A.equals( new StringBuilder(A).reverse().toString()) 
    ? "Yes" : "No" );

    int i=0,j=A.length()-1;
    while(A.charAt(i)==A.charAt(j) && (i++)<=(j--));
    System.out.println(i>=j?"Yes":"No");

Simple logic and would work even in C/C++

Most of the solutions posted here do a at least O(N). This isn't necessary, nor is making a reverse copy or char array. My code's preformance is O(N/2)

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();
        int lengthOfA = A.length()-1;
        String answer = "Yes";
        for(int i =0; i<lengthOfA/2; i++){
            if(A.charAt(i)==A.charAt(lengthOfA-i))
                continue;
            else{
                answer ="No";
                break;
            }
        }
        System.out.println(answer);
    }
}

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. */
    int len=A.length();
    char[] str=A.toCharArray();
    int flag=1;
    for(int i=0;i<len/2;i++){
        if(str[i]!=str[len-i-1]){
        System.out.println("No");
            flag=0;
        break;
    }
    }
    if(flag==1)
    System.out.println("Yes");
}

}

It passed all test cases

public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        String A=sc.next();
        int i = 0, j=A.length()-1;
        while(A.charAt(i)==A.charAt(j) && i++<j--);
        System.out.print(i>j?"Yes":"No");
    }