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].

题目：给定一个长度为2n(偶数)的数组，分成n个小组，返回每组中较小值的和sum，使sum尽量大

这道题让我们分割数组，两两一对，让每对中较小的数的和最大。这题难度不大，用贪婪算法就可以了。由于我们要最大化每对中的较小值之和，那么肯定是每对中两个数字大小越接近越好，因为如果差距过大，而我们只取较小的数字，那么大数字就浪费掉了。明白了这一点，我们只需要给数组排个序，然后按顺序的每两个就是一对，我们取出每对中的第一个数即为较小值累加起来即可，也就是说将数组从小到大排序，取下标为偶数的元素求和。

class Solution(object):
    def arrayPairSum(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return sum(sorted(nums)[::2])