338 Counting Bits

338. Counting Bits

1. Question

Given a non negative integer numbernum. For every numbersiin the range0 ≤ i ≤ numcalculate the number of 1's in their binary representation and return them as an array.

Example 1:

Input: 2
Output: [0,1,1]

Example 2:

Input: 5
Output: [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.

2. Implementation

(1) DP

思路: 通过观察，每个数的1bit个数等于这个数除以2的数的1bit个数，如果这个数是奇数的话，再这基础上再加上1

class Solution {
    public int[] countBits(int num) {
        int[] dp = new int[num + 1];

        for (int i = 1; i <= num; i++) {
            dp[i] = dp[i / 2] + (i & 1);
        }
        return dp;
    }
}

3. Time & Space Complexity

DP：时间复杂度O(n), 空间复杂度O(n)