原题：

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.

解析：

数组分割1

给一组2n个整数，分割为n组，一组2个数，使min(a,b)之和最大

Example:

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).

我的解法

很明显每组数值越接近，浪费的数值越小，所得的总和最大，我的解法如下

public class ArrayPartitionI {

    public static int arrayPartitionI(int[] array) {

        Arrays.sort(array);

        int sum = 0;

        for (int i = 0; i < array.length - 1; i += 2) {

            sum += array[i];

        }

        return sum;

    }

}

最优解法

public 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;

    }

}

哈哈几乎一样，后来想了一下不用length-1也可以的