10K's

Question

Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1's in the binary representation of i.

Algorithm

This question is actually asking us how do you count the 1's in an effective way.

By utilizing n&n-1 , we could zero out the least significant nonzero bit. Nothing to discuss, just a trick.

Code

class Solution {
    public int[] countBits(int n) {
        int[] ans = new int[n + 1];
        for (int i = 0; i <= n; i++) {
            ans[i] = helper(i);
        }
        return ans;
    }
    
    private int helper(int n) {
        int count = 0;
        while (n != 0) {
            count++;
            n &= n-1; // zeroing out the least significant nonzero bit
        }
        return count;
    }
}