README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0100-0199/0191.Number%20of%201%20Bits/README_EN.md	Bit Manipulation Divide and Conquer

191. Number of 1 Bits

中文文档

Description

Write a function that takes the binary representation of a positive integer and returns the number of set bits it has (also known as the Hamming weight).

 

Example 1:

Input: n = 11

Output: 3

Explanation:

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

Example 2:

Input: n = 128

Output: 1

Explanation:

The input binary string 10000000 has a total of one set bit.

Example 3:

Input: n = 2147483645

Output: 30

Explanation:

The input binary string 1111111111111111111111111111101 has a total of thirty set bits.

 

Constraints:

• 1 <= n <= 231 - 1

 

Follow up: If this function is called many times, how would you optimize it?

Solutions

Solution 1

Python3

class Solution:
    def hammingWeight(self, n: int) -> int:
        ans = 0
        while n:
            n &= n - 1
            ans += 1
        return ans

Java

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int ans = 0;
        while (n != 0) {
            n &= n - 1;
            ++ans;
        }
        return ans;
    }
}

C++

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int ans = 0;
        while (n) {
            n &= n - 1;
            ++ans;
        }
        return ans;
    }
};

Go

func hammingWeight(num uint32) int {
	ans := 0
	for num != 0 {
		num &= num - 1
		ans++
	}
	return ans
}

TypeScript

function hammingWeight(n: number): number {
    let ans: number = 0;
    while (n !== 0) {
        ans++;
        n &= n - 1;
    }
    return ans;
}

Rust

impl Solution {
    pub fn hammingWeight(n: u32) -> i32 {
        n.count_ones() as i32
    }
}

JavaScript

/**
 * @param {number} n - a positive integer
 * @return {number}
 */
var hammingWeight = function (n) {
    let ans = 0;
    while (n) {
        n &= n - 1;
        ++ans;
    }
    return ans;
};

C

int hammingWeight(uint32_t n) {
    int ans = 0;
    while (n) {
        n &= n - 1;
        ans++;
    }
    return ans;
}

Solution 2

Python3

class Solution:
    def hammingWeight(self, n: int) -> int:
        ans = 0
        while n:
            n -= n & -n
            ans += 1
        return ans

Java

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int ans = 0;
        while (n != 0) {
            n -= (n & -n);
            ++ans;
        }
        return ans;
    }
}

C++

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int ans = 0;
        while (n) {
            n -= (n & -n);
            ++ans;
        }
        return ans;
    }
};

Go

func hammingWeight(num uint32) int {
	ans := 0
	for num != 0 {
		num -= (num & -num)
		ans++
	}
	return ans
}

Rust

impl Solution {
    pub fn hammingWeight(mut n: u32) -> i32 {
        let mut res = 0;
        while n != 0 {
            n &= n - 1;
            res += 1;
        }
        res
    }
}