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.

題意：給定一個正整數，給出所有小於等於他的數的二進制中1的個數
思路：每一位都判斷，但是這不是O（n）的方法，看了看討論區，又一次感受到了智商的壓制。
O（n)的方法是利用已經求解的結果。一個數除2，就相當於二進制右移一位，所以n的二進制表示中1的個數就爲n/2的個數加上n最右一位的值，即ans[n] = ans[n/2]+ans[n]&0x01;
坑：暫無

代碼：

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> ans;
        for(int i = 0;i<=num;i++)
        {
            ans.push_back(getBitsNum(i));
        }
        return ans;
    }

    int getBitsNum(int num)
    {
        int t = 0;
        while(num!=0)
        {
            if(num&0x01) t++;

            num = num>>1;
        }
        return t;
    }
};

O（n）解法：

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