LeetCode题解——Counting Bits

题目：

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.

Hint:

1. You should make use of what you have produced already.
2. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
3. Or does the odd/even status of the number help you in calculating the number of 1s?

解题思路：

因为 i&(i-1) 的结果是将i的二进制表示中最右边的‘1’消除掉，因此i中1的个数和i&(i-1)中1的个数相差1.

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> res(num+1,0);
        for(int i=1; i<=num; i++){
            res[i] = res[i&(i-1)]+1;
        }
        return res;
    }
};