561. Array Partition I

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本文探讨了如何从一组整数中找到最佳配对，使得所有配对中较小数的总和达到最大。通过排序和选择策略，文章提供了一种简单而有效的解决方案，适用于竞赛编程和算法设计。

Easy

题目

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].

代码

python

思路：直接给整个list排序，然后按步长为2遍历整个list，选取最小的，求和即可。

class Solution:
    def arrayPairSum(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        nums=sorted(nums)
        sum=0
        for i in nums[::2]:#按步长为2遍历
            sum+=i
        return sum