leetcode:Counting Bits

原题：
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.

题意为：给定一个非负的整数num，对于0<=i<=num的i，计算i对应二进制数中1的个数，并以数组的形式返回

Java 实现：

public class Solution {    public int[] countBits(int num) {        int[] result = new int[num+1];        for (int i = 0; i <= num; i++) {            int temp= i;            int c=0;            String s = Integer.toBinaryString(temp);            for (int j = 0; j < s.length(); j++) {                if(s.charAt(j)=='1') c++;            }            result[i]=c;        }        return result;    }}