461. Hamming Distance (Easy)

TheHamming distancebetween two integers is the number of positions at which the corresponding bits are different.

Given two integersxandy, calculate the Hamming distance.

Note:
0 ≤x,y< 2^31.

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 1: Bit Maniulation O(1); O(1)

    public int hammingDistance(int x, int y) {
        int xor = x ^ y;
        int res = 0;
        while (xor != 0) {
            res += xor & 1;
            xor >>>= 1; //因为输入是非负数，所以可以不用无符号右移>>>，用>>即可
        }
        return res;
    }

Solution 2: Bit Maniulation O(1); O(1) 低效，不推荐，但是477要用到这种形式

    public int hammingDistance(int x, int y) {
        int count = 0;
        for (int i = 0; i < 32; i++) {
            if (((x % 2) ^ (y % 2)) == 1) { // the modulus is 2 instead of 10.
                count++;
            }
            x >>>= 1; //因为输入是非负数，所以可以不用无符号右移>>>，用>>即可
            y >>>= 1;
        }
        return count;
    }