首页 > 其他 > 详细

LeetCode 191. Number of 1 Bits

时间:2017-03-08 10:56:10      阅读:178      评论:0      收藏:0      [点我收藏+]

191. Number of 1 Bits

Description Submission Solutions

  • Total Accepted: 137583
  • Total Submissions: 354498
  • Difficulty: Easy
  • Contributors: Admin

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.

Subscribe to see which companies asked this question.

【题目分析】
给定一个无符号数,计算该数的二进制形式中包含的1的位数。
 
【思路】
用无符号右移动
public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int count = 0;
        
        while(n != 0) {
            count += n & 1;
            n = n >>> 1;
        }
        
        return count;
    }
}

 

LeetCode 191. Number of 1 Bits

原文:http://www.cnblogs.com/liujinhong/p/6517867.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!