首页 > 其他 > 详细

(easy)LeetCode 191.Number of 1 Bits

时间:2015-07-27 14:46:26      阅读:166      评论:0      收藏:0      [点我收藏+]

Number of 1 Bits

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.

参考编程之美120页

方法1:使用位操作

代码如下:

public class Solution {
// you need to treat n as an unsigned value
public int hammingWeight(int n) {
int num=0;
while(n!=0){
num+=(n&1);
n=n>>>1;
}
return num;
}
}

运行结果:时间复杂度为O(logV),即二进制位数。

技术分享

方法二:n&=(n-1)

代码如下:

public class Solution {
// you need to treat n as an unsigned value
public int hammingWeight(int n) {
int num=0;
while(n!=0){
n&=(n-1);
num++;
}
return num;
}
}

运行结果:

技术分享

 

 

(easy)LeetCode 191.Number of 1 Bits

原文:http://www.cnblogs.com/mlz-2019/p/4679860.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!