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leetcode------Number of 1 Bits

时间:2015-03-11 21:28:06      阅读:232      评论:0      收藏:0      [点我收藏+]
标题: Number of 1 Bits
通过率: 40.0%
难度:  简单

Write a function that takes an unsigned integer and returns the number of ’1‘ bits it has (also known as the Hamming weight).

For example, the 32-bit integer ’11‘ has binary representation 00000000000000000000000000001011, so the function should return 3.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

 

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本题就是数给的十进制数的二进制数有几个1。
最基本的做法就是32位扫描一遍进行查,我查看网页后找到一种简单的方法,减1法,每次把二进制减一,一直减到0为止,
n=n&(n-1);
具体代码如下:
 1 public class Solution {
 2     // you need to treat n as an unsigned value
 3     public int hammingWeight(int n) {
 4         int res=0;
 5         while(n!=0){
 6             n=n&(n-1);
 7             res++;
 8         }
 9         return res;
10     }
11 }

 

leetcode------Number of 1 Bits

原文:http://www.cnblogs.com/pkuYang/p/4330882.html

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