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@Yano 2017-12-10T02:05:34.000000Z 字数 569 阅读 1589

LeetCode 461 Hamming Distance

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题目描述

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 < 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.

分析

题目可以理解为:求两个数有多少位是不一样的。那么步骤可以为:

  1. 将两个数异或,结果记为 val
  2. 计算 val 有多少个 1

代码

  1. public int hammingDistance(int x, int y) {
  2. int val = x ^ y;
  3. int result = 0;
  4. while (val != 0) {
  5. if ((val & 1) == 1) {
  6. result++;
  7. }
  8. val >>= 1;
  9. }
  10. return result;
  11. }
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