python--leetcode561. 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].

这题的意思大概就是给2n个数字，按照（a1,b1）的方式组成n个序列，然后把所有min(ai，bi)加起来得合sum，求sum的最大值。

上代码：

class Solution(object):    def arrayPairSum(self, nums):        """        :type nums: List[int]        :rtype: int        """        nums.sort()        sum=0        for i in range(len(nums)):            if(i%2==0):  sum=sum+nums[i]        return sum

思路很简单，给list排个序，所有下标为偶数的元素加起来就为所求。

设想给定情况为[1,2,4,3]四个元素，组合成[1,2]、[3,4]这样所求sum值最大。

当然还有更简单的一行代码：

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