LeetCode No338. Counting Bits

1. 题目描述

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. 先找波规率, 如下图所示，由于低位数相同，我们可以发现[2~3]的一的数量是[0~1] 的基础上+1， 同理[4~7] 的1的数量是[0~3]的基础上+1. 总结下来就是[2^n ~ 2^(n+1) - 1]的1的数量是在[0 ~ 2^n - 1]的基础上+1.

数值 二进制 1的个数 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. 代码

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. 参考资料

5. 使用GCC内建的__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;    }};