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 = 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].

思路

因为要在两个数中取最小的值， 所以应该尽量找相邻的两个数一组， 这样才不会浪费一个大的数值。
所以将数组排序， 找到两个中间大的那一个

复杂度

时间O(nlogn) 排序的复杂度
空间O(1)

代码

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