# 561 Array Partition I

加總起來要得到最大值，就是小的跟小的在一起，大的跟大的在一起

題目連結在此

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.

input: 給一個偶數的陣列
output: 回傳 min(a1, b1) + ... min(ai, bi) 的最大值
這種中文好難翻譯，先看範例會比較清楚

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].
*/

/**
 * @param {number[]} nums
 * @return {number}
 */
var arrayPairSum = function(nums) {
  
};

Edge Case

• 有排序嗎？看起來是沒有

哪種資料結構解

一樣是 Array，然後會用到 Math.min()。我知道加總起來要得到最大值，就是小的跟小的在一起，大的跟大的在一起。所以首先要先排序 (很多題目都要先排序再做)，然後依序加起來。

var arrayPairSum = function(nums) {
    const len = nums.length;
    let result = 0;
    nums = nums.sort((a,b) => a-b);
    for(let i= 0; i < len; i += 2){
        result += Math.min(nums[i], nums[i+1])
    }

    return result;
};