leetcode 191. Number of 1 Bits(bit中1的个数)

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Write a function that takes an unsigned integer and returns the number of ‘1’ bits it has (also known as the Hamming weight).

Note:

Note that in some languages, such as Java, there is no unsigned integer type. In this case, the input will be given as a signed integer type. It should not affect your implementation, as the integer’s internal binary representation is the same, whether it is signed or unsigned.
In Java, the compiler represents the signed integers using 2’s complement notation. Therefore, in Example 3, the input represents the signed integer. -3.

Example 1:

Input: n = 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three ‘1’ bits.
Example 2:

Input: n = 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one ‘1’ bit.

Hamming距离,input是32位整型,统计其中bit位为1的个数。

思路:
有个著名的并行统计bit为1的算法,
先是相邻2位,然后4位,8位,具体参考并行bit count算法

    public int hammingWeight(int n) {
        n = n - ((n >> 1) & 0x55555555);               
        n = (n & 0x33333333) + ((n >> 2) & 0x33333333); 
        int c = ((n + (n >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
        
        return c;
    }
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今天发现一个面向程序员笔试的OJ(https://leetcode.com/problemset/algorithms/),上面的题目还算比较经典。感觉自己的算法并不是很好,所以从简单的题目开始练手。 191-Number of 1 Bits. 大意:给一个32-bit的整数,求二进制表示时1的个数。 解析:这道题是编程之美中的一道题。思路比较简单,直接填代码: int hammi

z159liao的专栏 387

Leetcode#191. Number of 1 Bits(位1的个数)

Leetcode#191. Number of 1 Bits(位1的个数) 题目描述 编写一个函数,输入是一个无符号整数,返回其二进制表达式中数字位数为 ‘1’ 的个数(也被称为汉明重量)。 示例 : 输入: 11 输出: 3 解释: 整数 11 的二进制表示为 00000000000000000000000000001011 示例 2: 输入: ...

武培轩 402

LeetCode191 Number of 1 Bits

Write a function that takes an unsigned integer and returnthe number of '1'bits it has (also known as theHamming weight). Example 1: Input: 00000000000000000000000000001011 Output: 3 Explanat...

李歇特冯-兹拜因巴哈的博客 188

Leetcode 191 Number of 1 Bits

Leetcode 191 Number of 1 Bits#include <stdint.h> using namespace std; class Solution { public: int hammingWeight(uint32_t n) { //n减1是将n最右边的那个1变成0,如果这个1后面跟着0,则这些0全变成1,相与可以得到最右边的1 //1

自己的学习笔记 243

LeetCode //C - 191. Number of 1 Bits

【代码】LeetCode //C - 191. Number of 1 Bits。

Made in Code 356

LeetCode:191. Number of 1 Bits位1的个数(C语言)

题目描述: 编写一个函数,输入是一个无符号整数,返回其二进制表达式中数字位数为 ‘1’ 的个数(也被称为汉明重量)。 示例 1: 输入:00000000000000000000000000001011 输出:3 解释:输入的二进制串 00000000000000000000000000001011 中,共有三位为 ‘1’。 示例 2: 输入:00000000000000000000000010000000 输出:1 解释:输入的二进制串 00000000000000000000000010000000 中,

C/C++攻城狮 309
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