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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?

Companies: Apple, Box, Amazon, Facebook, Qualcomm, Microsoft, Adobe, Cisco, Bloomberg, Google, Intel, Mindtickle

Related Topics:
Divide and Conquer, Bit Manipulation

Similar Questions:

Solution 1.

// OJ: https://leetcode.com/problems/number-of-1-bits/
// Author: github.com/lzl124631x
// Time: O(1)
// Space: O(1)
class Solution {
public:
    int hammingWeight(uint32_t n) {
        int ans = 0;
        for (; n; n -= (n & -n)) ++ans;
        return ans;
    }
};

Or

// OJ: https://leetcode.com/problems/number-of-1-bits/
// Author: github.com/lzl124631x
// Time: O(1)
// Space: O(1)
class Solution {
public:
    int hammingWeight(uint32_t n) {
        int ans = 0;
        for (; n; n = n & (n - 1)) ++ans;
        return ans;
    }
};

Or return __builtin_popcount(n);