Counting Bits

题目描述：

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.

如果用O(n*sizeof(integer))时间复杂度来做就太容易了。

如果用时间复杂度为O(n)的来写的话，先找到这么一个规律。

例如5表示为101,它的1的个数就是001(即5%4=1)的个数+1，6表示为110，它的1的个数就是010(即6%4=2)的个数+1.所以关键找到这个4，然后取余加1即可。

代码如下：

public class Solution {
    public int[] countBits(int num) {
        int[] dp=new int[num+1];
        dp[0]=0;
        int n=0;
        for(int i=1;i<=num;i++){
        	if(i==1<<(n+1))
        		n++;
        	dp[i]=dp[i%(1<<n)]+1;
        }
        return dp;
    }
}