561. Array Partition I

561. Array Partition I

难度: Easy

刷题内容

原题连接

内容描述

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:
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].

解题方案

思路 1
- 时间复杂度: O(NlgN)- 空间复杂度: O(1)

这个就是脑子一想就知道了,想要达到结果,我要尽可能取大一点,那么就是排好序每次取一对中的前一个就行了

例如[1,4,3,2]我们排好序为[1,2,3,4],然后让我们的pair为[1,2],[3,4],这样我们结果就是 1 + 3 = 4,肯定最大

beats 99.32%

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

思路 2

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