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:
1 class Solution { 2 public int[] countBits(int num) { 3 int [] res = new int[num + 1]; 4 for(int i = 1; i <= num; i++){ 5 res[i] = (i & 1) + res[i >> 1]; 6 } 7 8 return res; 9 } 10 }