Leetcode 561. Array Partition I（Easy）

1.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: 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].
翻译：给定2n个整数，你的任务是把这些整数组织成n对，例如（a1,b1）,(a2,b2)，...,(an,bn)使得1-n对儿中每对数的最小值的和最大。

2.思路

把这些数据由小到大进行排序，相邻两两一组，将每组中较小的数字相加求和。

3.算法

public int arrayPairSum(int[] nums) {
Arrays.sort(nums);
int res=0;
for(int i=0;i<nums.length;i=i+2){
res+=nums[i];
}
return res;
}

4.总结

也是比较简单的一道题，数组排序为Arrays.sort(nums);该方法返回为void.