【Leetcode】Counting Bits

题目链接：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.

思路：

c[i] ：数字i的二进制有多少个1 状态转移方程：

c[i]=c[i>>1]+i&1

比如数字7的二进制111只需要计算最后一位和前两位中1的个数，而前两位的个数和110的个数是一样的，也和11是一样的。

算法：

public int[] countBits(int num) {
int res[] = new int[num+1];
for(int i=0;i<=num;i++){
res[i] = res[i>>1]+(i&1);
}
return res;
}