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: 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：

求两两对最小值的最大值，排序以后每隔二取值即可。

class Solution {
public:
int arrayPairSum(vector<int>& nums) {
int sum=0;

sort(nums.begin(),nums.begin()+nums.size());

for(int i=0;i<nums.size();i=i+2){
sum+=nums[i];
}
return sum;
}
};

561. Array Partition I

原文：https://www.cnblogs.com/liangyc/p/8782155.html