[关闭]
@Yano 2017-12-10T02:05:34.000000Z 字数 569 阅读 936

LeetCode 461 Hamming Distance

LeetCode


LeetCode 排列组合 题目汇总
LeetCode 数字 题目汇总
LeetCode 动态规划 题目分类汇总
干货!LeetCode 题解汇总

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. }
添加新批注
在作者公开此批注前,只有你和作者可见。
回复批注