461. Hamming Distance

The 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.

class Solution {
    public int hammingDistance(int x, int y) {
        String xx = Integer.toBinaryString(x);
        String yy = Integer.toBinaryString(y);
        int res = 0;
        System.out.println(xx+" " + yy);
        if(xx.length() > yy.length()){
            int d = xx.length() - yy.length();
            while(d > 0){
                yy = "0" + yy;
                d--;
            }
        }
        if(yy.length() > xx.length()){
            int d = yy.length() - xx.length();
            while(d > 0){
                xx = "0" + xx;
                d--;
            }
        }
        for(int i = 0; i < xx.length(); i++){
            if(xx.charAt(i) != yy.charAt(i)) res++;
        }
        return res;
    }
}

不够的前面补0

class Solution {
    public int hammingDistance(int x, int y) {
        return Integer.bitCount(x^y);
    }
}

大佬的做法。。这就是差距吧，，比不同用异或，有多少位不同就有多少个1，然后bitCount完事

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来源： https://www.cnblogs.com/wentiliangkaihua/p/12886570.html