【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: 4
Explanation: 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].

題目的大體意思呢就是給一個數組，對立面的數字兩兩配對，然後取配對中最小的數求和，求這個和最大是多少。

思路也挺簡單的，對數組先排序，然後取所有的arr[2n]相加即可。

這裏需要注意的是：js中自帶的sort，對負數的排序有誤。比如[2，-1，-5,6].sort（）的結果是[-1，-5,2,6]，所以要自己寫排序

/**
 * @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;
};