LeetCode[561]Array Partition I

Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.

Example 1:

Input: [1,4,3,2]Output: 4Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).

Note:

1. n is a positive integer, which is in the range of [1, 10000].
2. All the integers in the array will be in the range of [-10000, 10000].

本地执行代码：

#include<iostream>#include<vector>#include<algorithm>using namespace std;/*解题思路：遍历向量，每次取出相邻的两个元素，然后比大小，对小的求和*/int arrayPairSum(vector<int>& nums);int getmin(int x, int y);int main(){vector<int> a(10, 1);int b = 0;b = arrayPairSum(a);cout << b << endl;system("pause");return 0;}int arrayPairSum(vector<int>& nums){int sum = 0;sort(nums.begin(), nums.end());for (int i = 0; i<int(nums.size()); i=i+2){sum = sum + getmin(nums[i], nums[i + 1]);}return sum;}int getmin(int x, int y){if (x > y){return y;}else{return x;}}

提交代码：

class Solution {public:    int getmin(int x, int y)    {        if (x > y)        {            return y;        }        else        {            return x;        }    }    int arrayPairSum(vector<int>& nums)     {        int sum = 0;        sort(nums.begin(), nums.end());        for (int i = 0; i<int(nums.size()); i=i+2)        {            sum = sum + getmin(nums[i], nums[i + 1]);        }        return sum;    }};