LeetCode No338. Counting Bits

1. 題目描述

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:

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

2. 思路

1. 先找波規率, 如下圖所示，由於低位數相同，我們可以發現[2~3]的一的數量是[0~1] 的基礎上+1， 同理[4~7] 的1的數量是[0~3]的基礎上+1. 總結下來就是[2^n ~ 2^(n+1) - 1]的1的數量是在[0 ~ 2^n - 1]的基礎上+1.

數值	二進制	1的個數
0	0	0
1	1	1
2	10	0 + 1
3	11	1 + 1
4	100	0 + 1
5	101	1 + 1
6	110	1 + 1
7	111	2 + 1
8	1000	0 + 1
9	1001	1 + 1
10	1010	1 + 1
11	1011	2 + 1
12	1100	1 + 1
13	1101	2 + 1
14	1110	2 + 1
15	1111	3 + 1

3. 代碼

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> count(num + 1); 
        count[1] = 1;

        for (int flag = 2, i = 2; i <= num && i < (flag << 1); ) {
            count[i] = count[i - flag] + 1;
            if (++i == flag << 1) 
                flag <<= 1;
        }
        return count;
    }
};

4. 參考資料

5. 使用GCC內建的__bulitin__popcount()

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