码迷,mamicode.com
首页 > 其他好文 > 详细

Hamming Distance

时间:2018-03-29 21:18:15      阅读:156      评论:0      收藏:0      [点我收藏+]

标签:blank   nbsp   nat   while   xpl   gdi   lan   target   solution   

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 ≤ xy < 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.

这种题真的会有各种技巧,先上个自己写的:
 1 class Solution {
 2 public:
 3     int hammingDistance(int x, int y) 
 4     {
 5         int z = x ^ y;
 6         int res = 0;
 7         while (z != 0)
 8         {
 9             int mod = z % 2;
10             res += mod;
11             z >>= 1;
12         }
13         return res;
14     }
15 };

贴一个优化过的:

 1 class Solution {
 2 public:
 3     int hammingDistance(int x, int y) {
 4         int z = x^y;
 5         int count = 0;
 6         while(z){
 7             count += z&1;
 8             z >>= 1;
 9         }
10         return count;
11     }
12 };

 

Hamming Distance

标签:blank   nbsp   nat   while   xpl   gdi   lan   target   solution   

原文地址:https://www.cnblogs.com/jiadyang/p/8671986.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!