LeetCode 461. Hamming Distance (汉明距离)

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.

 

 


 题目标签:Bit Manipulation
  这道题目给了我们两个int , 让我们找出hamming distance。设一个for loop 循环32次,然后每一次循环,利用 & 1 来取得最后一个bit, 比较x和y是不是相等,不相等的话,增加res的值by 1。再利用 >> 1来移动bits向右一位。
 

Java Solution:

Runtime beats 57.75% 

完成日期:06/28/2017

关键词:Bit Manipulation

关键点:利用 & 1拿到bit, 利用 >> 来移动bits

 

 1 public class Solution 
 2 {
 3     public int hammingDistance(int x, int y) 
 4     {
 5         int res = 0;
 6         
 7         for(int i=0; i<32; i++)
 8         {
 9             if((x & 1) != (y & 1))
10                 res++;
11             
12             x = x >> 1;
13             y = y >> 1;
14         }
15         
16         return res;
17     }
18 }

参考资料:N/A

 

LeetCode 算法题目列表 - LeetCode Algorithms Questions List

posted @ 2017-06-29 04:02  Jimmy_Cheng  阅读(318)  评论(0编辑  收藏  举报