561. Array Partition I

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本文介绍了一种通过排序并选择每对最小值来最大化配对和的算法。以2n个整数为例，将这些整数分成n对，使得每对中的最小值之和尽可能大。文章提供了一个具体的示例和简洁的Java实现。

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.

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].

题意：

把2n个数分成n组，一组两个数，求每组中较小数的和，使此时越大越好

算法思路：

把数组元素从小到大排序，输出奇数项的和即可

代码

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