LeetCode题目：338. Counting Bits

最新推荐文章于 2019-03-06 21:06:57 发布

原创最新推荐文章于 2019-03-06 21:06:57 发布 · 230 阅读

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本文介绍了一种计算从0到给定非负整数范围内每个数字二进制表示中1的数量的方法，并通过一个具体实例展示了算法实现，即对于输入5返回[0,1,1,2,1,2]。

题目原址：点击打开链接

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.

我的代码：

时间复杂度应该是O（num*log(num))

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> result;
        for(int i=0;i<=num;i++){
            int k=i;
            int sum=0;
            while(k){
                int j;
                j=k%2;
                k=k/2;
                if(j==1)sum++;
            }
            result.push_back(sum);
        }
        return result;
    }
};