leetCode 338. Counting Bits | Dynamic Programming | Medium

标签：dynamic programming

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].

题目大意：

给一个数字，比如5，那么5之前所有的整数的每个二进制表示中1的个数。

思路：

代码如下：

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> result;
        if(num == 0)
        {
            result.push_back(0);
            return result;
        }
        if(num == 1)
        {
            result.push_back(0);
            result.push_back(1);
            return result;
        }
        result.push_back(0);
        result.push_back(1);
        
        int temp = 2;
        for(int i = 2; i <=num ; i++)
        {
            if(i == temp*2)
                temp *= 2;
            result.push_back(result[i-temp] + 1);
        }
        return result;
    }
};

2016-09-01 18:57:26

本文出自 “做最好的自己” 博客，请务必保留此出处http://qiaopeng688.blog.51cto.com/3572484/1845319

leetCode 338. Counting Bits | Dynamic Programming | Medium

标签：dynamic programming

原文地址：http://qiaopeng688.blog.51cto.com/3572484/1845319