【LeetCode】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: 4Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).

Note:

1. n is a positive integer, which is in the range of [1, 10000].
2. All the integers in the array will be in the range of [-10000, 10000].

题目的大体意思呢就是给一个数组，对立面的数字两两配对，然后取配对中最小的数求和，求这个和最大是多少。

思路也挺简单的，对数组先排序，然后取所有的arr[2n]相加即可。

这里需要注意的是：js中自带的sort，对负数的排序有误。比如[2，-1，-5,6].sort（）的结果是[-1，-5,2,6]，所以要自己写排序

/** * @param {number[]} nums * @return {number} */function des(a,b){  return a-b;}var arrayPairSum = function(nums) {    nums.sort(des);    var ret=0;    for(var i =0;i<nums.length;i+=2){        ret+=nums[i];    }    return ret;};