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.

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].

public class Solution {
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;
}
}

一共有2n个元素，讲其两两组合，一共可以形成n对，找出n对中每对的最小值进行相加，求最大的和。
此时需要考虑如何保证抽取n的数能得到最大的sum。
将数组排序，按照从小到大的顺序进行排列
[a0,a1,a2,a3,a4……a2n-1]
两两组合，保证得到最大sum，则抽取的数为a1 a3 a5 a7.。。a2n-1
保证得到最小的sum,则抽取的数为a0 a1 a2..an-1