LeetCode-338. 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.

这道题主要需要一个先验知识就是 n=n&(n-1)会消去n最后面的一个1；然后就是将每一个数与之前数相与，比如b = n&(n-1)，那么n就比b多了一个1，而b比n小，b有几个1已知。

public class Solution {
public int[] countBits(int num) {
int[] temp = new int[num+1];
temp[0] = 0;
for(int i = 1; i <= num; i++){
temp[i] = 1 + temp[i&(i-1)];
}
return temp;

}
}