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::...
publicinthammingWeight(intn){intcount =0;for(inti=0; i<32; i++){if((n&1) ==1) { count++; } n = n >>1; }returncount; } 03 第二种解法 对于上面的解法,还可以再简化下,省掉那个判断,因为和1做与运算,如果最后一位是1,整个运算结果就是1,就相当于是自增加1。 publicinthammingWeight...
Write a function that takes an unsigned integer and return the number of '1' bits it has (also known as the Hamming weight). Example 1: Input: 00000000000000000000000000001011 Output: 3 Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits. Example 2...
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...
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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...
Basic idea: each time we take a look at the last four digits of binary verion of the input, and maps that to a hex char shift the input to the right by 4 bits, do it again until input becomes 0. https://discuss.leetcode.com/topic/60365/simple-java-solution-with-comment/2 ...
1866. 恰有 K 根木棍可以看到的排列数目 - 有 n 根长度互不相同的木棍,长度为从 1 到 n 的整数。请你将这些木棍排成一排,并满足从左侧 可以看到 恰好 k 根木棍。从左侧 可以看到 木棍的前提是这个木棍的 左侧 不存在比它 更长的 木棍。 * 例如,如果木棍排列为 [1,3,2,5
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. ...