题解 | #懂二进制#

思路：二进制不同位，直接采用异或，然后采用循环右移计算1的个数即可。

Java:

public class Solution {
    public int countBitDiff (int m, int n) {
        int c = m^n;
        int count = 0;
        while(c != 0){
            count += c & 1;
            c = c >> 1;
        }
        return count;
    }
}

C++:

class Solution {
public:
    int countBitDiff(int m, int n) {
        int count = 0;
        m = m ^ n;
        for(int i = 0; i < 32; ++i){
             count += (m >> i) & 1;
        }
        return count;
    }
};