191. Number of 1 Bits

Description

Write a function that takes the binary representation of 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.

Example 3:

Input: n = 11111111111111111111111111111101
Output: 31
Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.

 

Constraints:

  • The input must be a binary string of length 32.

 

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

Solutions

Solution 1

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

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

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

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

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

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

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

Solution 2

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

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

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

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