338. Counting Bits"

338. Counting Bits

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.

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Example 1:

Input: n = 2
Output: [0,1,1]
Explanation:
0 --> 0
1 --> 1
2 --> 10

Example 2:

Input: n = 5
Output: [0,1,1,2,1,2]
Explanation:
0 --> 0
1 --> 1
2 --> 10
3 --> 11
4 --> 100
5 --> 101

Constraints:

0 <= n <= 105

Follow up:

It is very easy to come up with a solution with a runtime of O(n log n). Can you do it in linear time O(n) and possibly in a single pass?
Can you do it without using any built-in function (i.e., like __builtin_popcount in C++)?

Solution

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class Solution {
    public int[] countBits(int n) {
        int[] bits = new int[n + 1];
        
        for (int num = 0; num < bits.length; num++) {
            int count = 0;
            int i = 0;
            while (i < 32) {
                if ((num & (1 << i)) != 0) {
                    count++;
                }
                i++;
            }
            bits[num] = count;
        }
        
        return bits;
    }
}

Solution#

Solution