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].
   
   

题目的要求是将所给数组划分成n对，取出每对整数中的最小值，依次相加求和，通过划分使得求得的和最大。

通过例子，我们不难发现，其实最佳的划分思路就是按照大小排序之后去划分，其实这种划分思路说到低用的是贪心的思想，让在相同位置上每次取得的min都尽可能得大，然后求和。这样代码的思路就十分清晰了：先排序，再求和：

class Solution {public:    int arrayPairSum(vector<int>& nums) {        int size = nums.size() / 2;        sort(nums.begin(), nums.end());        int sum = 0;        for (int i = 0 ; i < size; i++) {            int min = 1000000;            for (int j = 2 * i; j <= 2 * i + 1; j++) {                if (nums[j] < min)                    min = nums[j];            }            sum += min;        }        return sum;    }};

可以看到，我是先对数组排序，然后按照可以划分的对数，算出每一对的最小值，依次相加。事实上，在排完顺序之后，完全可以直接求和。因为我们是每一对去求最小值，在排完序之后，我们直接间隔取值就可以了：

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