C++实现LeetCode(191.位1的个数)

[LeetCode] 191.Number of 1 Bits 位1的个数

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.

很简单的一道位操作Bit Manipulation的题,最近新出的三道题都没有啥难度啊,这样会误导新人的,做了这三道得出个LeetCode没啥难度的结论,其实里面好题真的不少,难题也很多,经典题也多,反正就是赞赞赞,32个赞。

class Solution {public:    int hammingWeight(uint32_t n) {        int res = 0;        for (int i = 0; i < 32; ++i) {            res += (n & 1);            n = n >> 1;        }        return res;    }};

到此这篇关于C++实现LeetCode(191.位1的个数)的文章就介绍到这了,更多相关C++实现位1的个数内容请搜索以前的文章或继续浏览下面的相关文章希望大家以后多多支持!

阳光总在风雨后。只有坚强的忍耐顽强的奋斗,

C++实现LeetCode(191.位1的个数)

相关文章:

你感兴趣的文章:

标签云: