LeetCode 338.Counting Bits 题解（C++）

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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.

示例

• For num = 5 you should return [0,1,1,2,1,2]

要求

• 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.

思路

利用十进制转换二进制的算法，在转换过程中利用temp保存取余后得到1的个数，即为该数字二进制1的个数。特别需要注意的是，在while循环结束后，需要再进行一次取余的判断，否则会丢失一个二进制1.

代码

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

补充

从别人学习到的DP算法：

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