LeetCode 338. Counting Bits

原题链接在这里：https://leetcode.com/problems/counting-bits/

题目：

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array.

Example:
For num = 5 you should return [0,1,1,2,1,2].

Follow up:

• It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?
• Space complexity should be O(n).
• Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

题解：

Take an example: num, binary representation is 1101.

it contains two parts. The last digit, num & 1.

The other digits, 110, which has been calculated before. res[num >> 1].

Time Complexity: O(num).

Space: O(n), res array.

AC Java:

class Solution {
    public int[] countBits(int num) {
        int [] res = new int[num + 1];
        for(int i = 1; i <= num; i++){
            res[i] = (i & 1) + res[i >> 1];
        }
        
        return res;
    }
}

类似Number of 1 Bits.