Level：
Medium
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 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.
思路分析：
我们可以用动态规划的思想来解决，dp[ i ]，表示，i的二进制中1的个数，那么状态转移方程为dp[ i ]=dp[i>>1]+(i%2)。
代码：
public class Solution{
public int []countBits(int num){
int []dp=new int [num+1];
for(int i=0;i<=num;i++){
dp[i]=dp[i/2]+i&1;
}
return dp;
}
}