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tags: Array

从结果出发, 只需要找到加法的结果，而不强调具体配对。

找到排列取单数位的规律，再考虑负数和正数的相同规律，即可找到排列求解的方法。

```

/*

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).

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].

*/

/*

Thoughts: goal is to find the half of the numbers' sum, and always pick the min value of the pair.

Also, need to make the overall sum as large as possible: can't always choose the smallest numbers, but we can choose numbers at ascending order.

1. sort array.

2. only pick the even ones (starting from index 0)

Note:

1. use long to save result: never know what sum can occur in the process.

2. sort the array

O(nlogn)

*/

class Solution {

public int arrayPairSum(int[] nums) {

if (nums == null || nums.length <= 1) {

return 0;

}

Arrays.sort(nums);

long result = 0;

for (int i = 0; i < nums.length; i++) {

result += i % 2 == 0 ? nums[i] : 0;

}

return (int)result;

}

}

```