561. Array Partition I

题目描述：

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: 4
Explanation: 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]

解题思路：

为了使和最大，这就要求每一组中的最小数尽量大，这要求小的数与比它更小的数组成一组。

将输入向量排序后，从最小的数开始，两两组成一组，此时得到的和最大。

代码：

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