1.題目描述

338. Counting Bits

Medium

101773

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 1:

Input: 2
Output: [0,1,1]

Example 2:

Input: 5
Output: [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.

2.解題思路

觀察可以發現一串數 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111.........可以通過前面的數在末尾加上0或1形成

可以把這些數寫成一個二叉樹結構

                                                                0

                                                       10             11

                                              100    101      110  111

                                                   .............................

可以看出  數字i中1的個數 = i的父親節點1的個數 + 1（若i是偶數）

用v[i]記數字i中1的個數， 上面的式子就可以寫成v[i] = v[i/2] + i%2

3.實現程式碼

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> V;
        V.push_back(0);
        for (int i=1; i<=num; i++) {
            V.push_back(V[i/2]+i%2);
        }
        return V;
    }
};