Leetcode 561. Array Partition I(easy)

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].

思路：

需要注意的是找出配对后最小值的和的最大值，其实就是要求最小值要与另一个数的差值最小，这样最小值才有机会获得更大的值。这样，让每个配对的数中较大值取为较小值的后一个数。

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