LeetCode 461. Hamming Distance

原题链接在这里：https://leetcode.com/problems/hamming-distance/

题目：

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.

题解：

Bits different自然而然就想到了xor.

利用Number of 1 Bits的数xor 结果的bit 1个数.

或者用Integer.bitCount() function.

Time Complexity: O(1). Space: O(1).

AC Java:

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

 跟上Total Hamming Distance.