LeetCode 191:number of one bits
题目就是:
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.
这就是考十进制转二进制的题,学到了#include <cstdin>这个头文件,还有简单的转换算法,算法里边数据类型,别随手定义成了int!!!
class Solution {
public:
int hammingWeight(uint32_t n)
{
uint32_t iRet = ;
uint32_t flag = ;
uint32_t iRes = n;
uint32_t iLeft = ;
while(flag != )
{
flag = iRes / ;
iLeft = iRes - flag * ;
if(iLeft == )
{
++iRet;
}
iRes = flag;
}
return iRet;
}
};
写的不好,但多code多学习总是有用。
LeetCode 191:number of one bits的更多相关文章
- LeetCode OJ:Number of 1 Bits(比特1的位数)
Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also know ...
- leetcode:Number of 1 Bits
Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also know ...
- LeetCode 762 Prime Number of Set Bits in Binary Representation 解题报告
题目要求 Given two integers L and R, find the count of numbers in the range [L, R] (inclusive) having a ...
- [LeetCode] 762. Prime Number of Set Bits in Binary Representation_Easy
Given two integers L and R, find the count of numbers in the range [L, R] (inclusive) having a prime ...
- LeetCode算法题-Number of 1 Bits(Java实现)
这是悦乐书的第186次更新,第188篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第45题(顺位题号是191).编写一个带无符号整数的函数,并返回它所具有的"1 ...
- LeetCode(476): Number Complement
Given a positive integer, output its complement number. The complement strategy is to flip the bits ...
- LeetCode OJ:Number of Islands(孤岛计数)
Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surro ...
- LeetCode 693 Binary Number with Alternating Bits 解题报告
题目要求 Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits w ...
- Leetcode 762. Prime Number of Set Bits in Binary Representation
思路:动态规划.注意1024*1024>10^6,所以质素范围是(0,23). class Solution { public int countPrimeSetBits(int L, int ...
随机推荐
- 使用Cydia Substrate Hook Android Java世界
从来没接触过Android的HOOK,在看雪上找到了一篇HOOK 的文章,但是太复杂了,应该是本地环境问题,测试不成功. 后来搜到Cydia Substrate,看了几篇文章,进入官网查看了一下文档, ...
- 系统学习Docker 践行DevOps理念
Docker代表的容器技术是近两年的大热技术,和人工智能.区块链等热点不同,容器技术的门槛并不高,每一个开发.测试.运维人员都能在日常工作中掌握和使用,是当今IT从业人员的必备技能之一.本课程会带大家 ...
- python安装Django
现在有很多建站系统,很多都是基于php的,比如WordPress. 而Django 是老牌基于Python的CMS框架了,一直听说很强大,甚至曾经很红的Ruby On Rails都参考了它的很多概念, ...
- java script 学习
用JavaScript输出文本 <p>我的第一个段落.</p> <script> document.write(Date()); </script> & ...
- 使用IMAGEMAGICK的CONVERT工具批量转换图片格式
使用IMAGEMAGICK的CONVERT工具批量转换图片格式 http://www.qiansw.com/linux-imagemagick-convert-img.html Home > 文 ...
- linux文件系统(ext2)
一个磁盘可以划分成多个分区,每个分区必须先用格式化工具(例如某种mkfs命令)格式化成某种格式的文件系统,然后才能存储文件,格式化的过程会在磁盘上写一些管理存储布局的信息.下图是一个磁盘分区格式化成e ...
- sessionStorage & URL Origin
sessionStorage & URL Origin same origin https://developer.mozilla.org/en-US/docs/Web/API/Window/ ...
- RunKit & NPM
RunKit + NPM Try any Node.js package right in your browser https://npm.runkit.com/segmentit
- [剑指Offer] 38.二叉树的深度
题目描述 输入一棵二叉树,求该树的深度.从根结点到叶结点依次经过的结点(含根.叶结点)形成树的一条路径,最长路径的长度为树的深度. [思路1]递归 /* struct TreeNode { int v ...
- Java IO 之 FileFilter与FilenameFilter
FileFilter与FilenameFilter可以实现对文件的过滤,他们都是接口,具体的过滤规则需要我们自己编写 1.FileFilter package org.zln.io.file; imp ...