leetcode 338. Counting Bits

最新推荐文章于 2020-01-28 02:59:38 发布

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本文介绍了一种高效计算从0到给定整数范围内所有数的二进制表示中1的数量的方法。通过发现二进制数的规律，实现了一个线性时间复杂度的算法，避免了使用内置函数。

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.

1、不可能挨个数去计算。换成二进制是有规律的。

class Solution {
public:  
    vector<int> countBits(int num)
    {
        vector<int> ret(1, 0);
        if (num == 0) return ret;
        ret.push_back(1);
        if (num == 1) return ret;
        int last_important_num = 1;
        for (int i = 2; i <= num; i++)
        {
            if (i == last_important_num * 2)    //说明是2的阶乘
            {
                ret.push_back(1);
                last_important_num = i;
                continue;
            }
            else
            {
                ret.push_back(1 + ret[i - last_important_num]);
            }
        }
        return ret;
    }
};