[LeetCode] Counting Bits

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.

public class Solution {    public int[] countBits(int num) {    int[] dp=new int[num+1];    for(int i=0;i<=num;i++){    dp[i]=get1s(i);    }    return dp;    }    public int get1s(int num){    int re=0;    for(int i=1;i!=0;i<<=1){    if((i&num)>0) re++;    }    return re;    }}

public class Solution2 {public int[] countBits(int num) {int[] dp=new int[num+1];dp[0]=0;for(int i=1;i<=num;i++){dp[i]=dp[i&i-1]+1;//这个技巧在位运算中特别常见 包括在树状数组的实现代码中}return dp;} }