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.思路： (1)先求x^y的结果res。 (2)再依次求32位res的每一位与1进行与操作的结果，若不为0，则Hamming Distance加一。 (3)每判断完一位，res右移一位继续判断下一位。
public class Solution { public int hammingDistance(int x, int y) { int res = x ^ y; int count = 0; for (int i = 0; i < 32; i++) { if ((res & 1) != 0) count++; res >>= 1; } return count; } }