LeetCode——Counting Bits

Question

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.
Credits:
Special thanks to @ syedee for adding this problem and creating all test cases.

Solution

动态规划，这里要应用求一个整数，对应二进制数中1的个数的方法，i & (i - 1)，也就是将i的最右边的1置为0.

那么每个整数中1的个数，就等于不包含最右边1以后1的个数再加1。 ret[i] = ret[i & (i - 1)] + 1.

Code

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