191. Number of 1 Bits

leetcode easy

#bit-manipulation#csharp#easy#leetcode#math#string

Task

Напишите функцию, которая принимает бинарное представление положительного целого числа и возвращает количество установленных битов (также известных как вес Хэмминга).

Пример:

Input: n = 11

Output: 3

Explanation:

The input binary string 1011 has a total of three set bits.

C# solution

matched/original

int hammingWeight(int n) {
    int bits = 0;
    int mask = 1;
    for (int i = 0; i < 32; i++) {
        if ((n & mask) != 0) {
            bits++;
        }
        mask <<= 1;
    }
    return bits;
}

C++ solution

auto-draft, review before submit

#include <bits/stdc++.h>
using namespace std;

// Auto-generated C++ draft from the C# solution. Review containers, LINQ and helper types before submit.
int hammingWeight(int n) {
    int bits = 0;
    int mask = 1;
    for (int i = 0; i < 32; i++) {
        if ((n & mask) != 0) {
            bits++;
        }
        mask <<= 1;
    }
    return bits;
}

Java solution

matched/original

public int hammingWeight(int n) {
    int bits = 0;
    int mask = 1;
    for (int i = 0; i < 32; i++) {
        if ((n & mask) != 0) {
            bits++;
        }
        mask <<= 1;
    }
    return bits;
}

JavaScript solution

matched/original

function hammingWeight(n) {
    let bits = 0;
    let mask = 1;
    for (let i = 0; i < 32; i++) {
        if ((n & mask) !== 0) {
            bits++;
        }
        mask <<= 1;
    }
    return bits;
}

Python solution

matched/original

def hammingWeight(n: int) -> int:
    bits = 0
    mask = 1
    for _ in range(32):
        if (n & mask) != 0:
            bits += 1
        mask <<= 1
    return bits

Go solution

matched/original

func hammingWeight(n int) int {
    bits := 0
    mask := 1
    for i := 0; i < 32; i++ {
        if (n & mask != 0 {
            bits++
        }
        mask <<= 1
    }
    return bits
}

Explanation

Algorithm

1️⃣