easy 1 : 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 = 4Output: 2Explanation:1   (0 0 0 1)4   (0 1 0 0)       ↑   ↑The above arrows point to positions where the corresponding bits are different.

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“汉明距离”，一个较为著名的问题，给定两个整数，求出这两个数字的二进制中位数不同的个数。比如上面的1和4，在第0位和第2位数字不同，因此这个汉明距离就是2。

public class Solution {
    public int hammingDistance(int x, int y) {
        int z = x ^ y;//异或
        int res = 0;
        while(z != 0) {
            if(z%2 == 1) {
                res ++;
            }
            z >>= 1;
        }
        return res;
    }
}

 

shawngao 

Reputation:  1,163

What does come to your mind first when you see this sentence "corresponding bits are different"? Yes, XOR! Also, do not forget there is a decent function Java provided: Integer.bitCount() ~~~

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