notes/OJ notes/pages/Leetcode Number-of-1-Bits.md

Leetcode Number-of-1-Bits

2022-07-22 14:45

Algorithms:

#algorithm #bit_manipulation

Data structures:

#DS #bitset

Difficulty:

#coding_problem #difficulty-easy

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

tag:#bit_manipulation

Links:

• Link to problem

---

Problem

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

Examples

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.

Thoughts

[!summary] This is a #bit_manipulation problem.

Two methods for this problem.

Method 1: cpp's STL implementation

simply use bitset::count

Method 2: (n & (n - 1)) method

By using n = (n & (n - 1)), we can remove the last true in the original bitset:

5 : 101
4 : 100
5 & 4 : 100

10 : 1010
9  : 1001
10 & 9 : 1000

n - 1 changes the trailing falses to true, and change the last true to false, and by AND operation, we can remove the last true bit.

Solution

CPP STL:

class Solution {
public:
  int hammingWeight(uint32_t n) { return bitset<32>(n).count(); }
};

Method 2:

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