题目描述：
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].
题目大意：
给定一个长度为2n的数组，要把它分成n个分组，即每组有两个数，返回每组中最小值的总和，使和最大。
理解了大意就知道思路了，又看了下论坛里的算法分析
解决方案基本就是先按从小到大排序，这样相邻的数字是最接近的，然后再分成两两一组，取每组中的第一个数相加即可。
package main
import (
"fmt"
"sort"
)
func main() {
nums := []int{4, 5, 6, 1}
n := arrayPairSum(nums)
fmt.Println(n)
}
func arrayPairSum(nums []int) int {
sort.Ints(nums)
sum := 0
length := len(nums)
for i := 0; i < length; i += 2 {
sum += nums[i]
}
return sum
}
在线查看结果:The Go Playground
其实这里主要用到了Go的sort包给int数组排序。排序后遍历数组，每次递增2就可以了。