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).

含义： 给定一个长度为2n(偶数)的数组，分成n个小组，每组中取一个较小值出来，计算所有较小值的总和。请合理分组，使得较小值的总和sum尽量大

思路：先排序，将相邻两个数分为一组，每组较小数都在左边，求和即可

    public int arrayPairSum(int[] nums) {
        Arrays.sort(nums);
        
        int sum = 0;
        for (int i=0;i<nums.length;i+=2)
        {
            sum += nums[i];
        }
        return sum;
    }