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::...
开始我的代码: publicinthammingWeight(intn) {intcount=0;while(n!=0) { count+=n&0x1; n=n>>1; }returncount; }
n += n >> 16; // put count of each 32 bits into those 8 bits return n & 0xFF; } // This uses fewer arithmetic operations than any other known implementation on machines with fast multiplication. // It uses 12 arithmetic operations, one of which is a multiply. int hammingWeight(uin...
https://leetcode.com/problems/number-of-1-bits/description/ 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:Theinput binarystring00000000000000000000000...
Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits. Example 2: Input: 00000000000000000000000010000000 Output: 1 Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit. ...
Can you solve this real interview question? Number of Digit One - Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n. Example 1: Input: n = 13 Output: 6 Example 2: Input: n = 0 O
int countDigitOne(int n) { // mulk 表示 10^k // 在下面的代码中,可以发现 k 并没有被直接使用到(都是使用 10^k) // 但为了让代码看起来更加直观,这里保留了 k long long mulk = 1; int ans = 0; for (int k = 0; n >= mulk; ++k) { ans += (n / (mulk * 10)) ...
【LeetCode & 剑指offer刷题】特殊数题1:43 1~n整数中1出现的次数 (233. Number of Digit One )...,程序员大本营,技术文章内容聚合第一站。
其实Leetcode这道题的高票答案已经说的很详细了,我自己也是研究了很久才想明白,分享给大家,喜欢看英文的同学直接点链接好了。 顺带一说,这些人的脑子真的太好了,感觉我们这种人就是弱智一般。。 答案在此 废话少说,直接上代码。 int countDigitOne(int n) { ...
Explanation:The input binary string00000000000000000000000010000000has a total of one '1' bit. Example 3: Input:11111111111111111111111111111101 Output:31 Explanation:The input binary string11111111111111111111111111111101has a total of thirty one '1' bits. ...