（Medium) Hamming Distance LeetCode

Description:

he Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, calculate the Hamming distance.

Note:
0 ≤ x, y < 2³¹.

Example:

Input: x = 1, y = 4

Output: 2

Explanation:
1   (0 0 0 1)
4   (0 1 0 0)
       ↑   ↑

The above arrows point to positions where the corresponding bits are different.

Solution 

class Solution {
    public int hammingDistance(int x, int y) {
        
        
        String A = GetBit(x);
        
        String B = GetBit(y);
        
        int a = A.length();
        
        int b = B.length();
        
        int bigger  = a>b? a:b;
        
        int smaller = a>b? b:a;
        
        int result = 0;
        
        if(a>b){
            
            for(int i = 0; i<smaller;i++){
                
                if(A.charAt(i)!= B.charAt(i)){
                    result++;
                }
            }
            
            for(int i = smaller;i<bigger; i++){
                
                if(A.charAt(i)!='0'){
                    result++;
            }
         }
        }
        else{
             for(int i = 0; i<smaller;i++){
                
                if(A.charAt(i)!= B.charAt(i)){
                    result++;
                }
            }
            
            for(int i = smaller;i<bigger; i++){
                
                if(B.charAt(i)!='0'){
                    result++;
            }
            
        }
      
        }
        return result;
    }
    
    
    public String GetBit(int N){
        
        
        String res = "";
        
        int remain = N;
        
        do {
          int bit = remain%2;
        
          remain = remain/2;
            
          res = res +bit;
        }
        
        while (remain>0);
        
        return res;
    }
    
    
    
}