LeetCode 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), …, (a_n, b_n) which makes sum of min(a_i, b_i) 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:

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].

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给定一个长度为2n的数组，要求把数组分成n组，即(a₁, b₁), (a₂, b₂), …, (a_n, b_n) ，使得每组的最小值之和最大。 使用贪心策略，比如样例中，4和3肯定不能和1配对，因为1肯定是最小的，不能拿很大的数和1陪葬，所以只拿2和1配对；然后3、4配对。所以规律就是对数组排序，从小到大每两个配对。那么每组最小值之和就是第0、2、4…个数之和。 代码如下： [cpp] class Solution { public: int arrayPairSum(vector<int>& nums) { sort(nums.begin(), nums.end()); int ans = 0; for(int i = 0; i < nums.size(); i += 2) ans += nums[i]; return ans; } }; [/cpp] 本代码提交AC，用时92MS。]]>