561. Array Partition I

iven 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).
从优化上来说 只要尽可能的把小的聚集在一起，把大的聚集在一起就好了。

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