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：排序数组，取相邻两个数最小值，步数2遍历

class Solution {
    public int arrayPairSum(int[] nums) {
        int sum=0;
        Arrays.sort(nums);
        for (int i = 1; i < nums.length; i+=2) {
            sum += Math.min(nums[i],nums[i-1]);
        }
        return sum;
    }
}