C++ Solution: n & (n - 1)


  • 57
    H
    int hammingWeight(uint32_t n) {
        int count = 0;
        
        while (n) {
            n &= (n - 1);
            count++;
        }
        
        return count;
    }
    

    n & (n - 1) drops the lowest set bit. It's a neat little bit trick.

    Let's use n = 00101100 as an example. This binary representation has three 1s.

    If n = 00101100, then n - 1 = 00101011, so n & (n - 1) = 00101100 & 00101011 = 00101000. Count = 1.

    If n = 00101000, then n - 1 = 00100111, so n & (n - 1) = 00101000 & 00100111 = 00100000. Count = 2.

    If n = 00100000, then n - 1 = 00011111, so n & (n - 1) = 00100000 & 00011111 = 00000000. Count = 3.

    n is now zero, so the while loop ends, and the final count (the numbers of set bits) is returned.


  • 4
    D

    Worth mentioning that this solution will only loop as many times as the number of 1 bits whereas the solutions shifting the bits of n will loop also for every 0 bit in between.


  • 0
    H

    Yes, thank you. This should have been explained, as this is the crux of the algorithm.


  • 0
    Z
    This post is deleted!

  • 0
    W
    This post is deleted!

  • 0
    E

    Thanks for the explanation!


Log in to reply
 

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.