LeetCode.561 Array Partition I

题目：

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: 4Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).

Note:

1. n is a positive integer, which is in the range of [1, 10000].
2. All the integers in the array will be in the range of [-10000, 10000].

分析：

class Solution {    public int arrayPairSum(int[] nums) {        //给定数组长度为2n的数组，将其分为n对元素组合，求这些组合最小元素的之和        //思路：对元素进行排序，将排序的元素分为n组，取每组的第一个元素求和        int len=nums.length/2;                Arrays.sort(nums);        int sum=0;        for(int i=0;i<nums.length;i+=2){            sum+=nums[i];        }        return sum;    }}