笔试题13. LeetCode OJ (13) 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.
Note:
n is a positive integer, which is in the range of [1, 10000].
All the integers in the array will be in the range of [-10000, 10000].

题目意思是：给你2n个数，把这2n个数两两分成一组，求“每组最小值之和”的最大值。

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

只需将给定序列排序，然后从小到大依次将两个数分成一组（这样分组可使得“每组最小值之和”最大），将每组中的较小的数累加即为所要求得数。