LeetCode题解之Number of 1 Bits 1、题目描述 2.问题分析 使用C++ 标准库的 bitset 类,将整数转换为 二进制,然后将二进制表示转换为字符串,统计字符串中 1 的个数即可。 3、代码 1inthammingWeight(uint32_t n) {2bitset<32>b(n);3stringb_s =b.to_string() ;45intcount_one =0;6for(string::...
1.Number of 1 Bits 本题收获: 1.Hamming weight:即二进制中1的个数 2.n &= (n-1)【n = n & (n-1)】的用处 题目: Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as theHamming weight). For example, the 32-bit integer ’11...
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. 2.解决方案1 AI检测...
classSolution{public:intrangeBitwiseAnd(intm,intn){intbitm=0,bitn=0;for(inti=0;i<31;i++){if(m&(1<<i))bitm=i;if(n&(1<<i))bitn=i;}if(bitm==bitn){intsum=m;for(inti=m;i<n;i++)// 为了防止 2147483647+1 超过范围sum=(sum&i);sum=(sum&n);returnsum;}elsereturn0;}}...
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[LeetCode] Sparse Matrix Multiplication 稀疏矩阵相乘 Given two sparse matrices A and B, return the result of AB. You may assume that A's column number is ... 537 Complex Number Multiplication 复数乘法 详见:https://leetcode.com/problems/complex-number-multiplication/description/ C++: class ...
Leetcode 191. Number of 1 Bits 题目描述:统计二进制中1的位数。 题目链接:191. Number of 1 Bits 初步想法就是沿用上题190的思路,右移n,然后不断按位与操作来进行计数。 当然也可以用我以前面试的时候答的转成字符串哈哈哈~ 参考链接 191. Number of 1 Bits | 汉明重量几种解法......
191. Number of 1 Bits 题目: https://leetcode.com/problems/number-of-1-bits/ 难度: Easy 转成二进制,数1的个数
1. Description Sort Integers by The Number of 1 Bits 2. Solution 解析:由于最大数字不超过10000,因此1的位数不超过14位,注意+=的运算优先级要低于&,而+的运算优先级要高于&。Version 1用右移运算获得1的个数,Version 2用的bin函数,Version 3通过python自带的排序函数进行排序。Version 4使用字典保存结果。
var countPrimeSetBits = function(L, R) { // 存储二进制中1的个数 let arr = [] for (let i = L; i <= R; i++) { arr.push(i.toString(2).replace(/0/g, '').length) } // 判断一个数是否为质数 let isPrime = function(num) { if (num === 1) { re...