561.数组合并I（javascript）561.ArrayPartitionI

给定长度 2n 整数数组 nums ，你的任务是划分这些数字 n 对, 例如 (a1, b1), (a2, b2), …, (an, bn) ，使得从 1 到 n 的 min(ai, bi) 总和是最大的。

返回该 最大总和 。

Given an integer array nums of 2n integers, group these integers into n pairs (a1, b1), (a2, b2), …, (an, bn) such that the sum of min(ai, bi) for all i is maximized. Return the maximized sum.

示例 1：

输入：nums = [1,4,3,2]
输出：4
说明：所有可能的分数方法（忽略元素的顺序）都是：
1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4
所以最大总和是 4

示例 2：

输入：nums = [6,2,6,5,1,2]
输出：9
说明：最佳分数法是 (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9

Example 1:

Input: nums = [1,4,3,2]
Output: 4
Explanation: All possible pairings (ignoring the ordering of elements) are:
1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4
So the maximum possible sum is 4.

Example 2:

Input: nums = [6,2,6,5,1,2]
Output: 9
Explanation: The optimal pairing is (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9.

提示：

• 1 <= n <= 104
• nums.length == 2 * n

从示例中可以看出该定律：

1. 升序排列

2. 取下标为0或偶数，所有数字的总和

   var arrayPairSum = function (nums) { nums.sort((a, b) => a - b) let sum = 0 nums.forEach((item, index) => { if (index == 0 || index % 2 == 0) { sum += item }

   });
   return sum

   };

   var arrayPairSum = function (nums) { nums.sort((a, b) => a - b) let sum = 0 let len=nums.length for(let i=0;i<len;i+=2){ sum += nums[i] } return sum };

leetcode： https://leetcode-cn.com/problems/array-partition-i/