LeetCode 561 Array Partition I

最新推荐文章于 2025-08-19 16:19:29 发布

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最新推荐文章于 2025-08-19 16:19:29 发布

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文章标签：数据结构与算法

原文链接：https://juejin.im/post/5a37792151882538650948fb

本文介绍了一种算法问题：给定一个包含2n个整数的数组，如何将这些整数分成n对，使得每一对中较小数的总和尽可能大。通过排序并选择每个相邻数对中的第一个数来实现这一目标。

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##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].

Subscribe to see which companies asked this question. ##代码

func arrayPairSum(_ nums: [Int]) -> Int {
    guard nums.count % 2 == 0 else {
        return 0
    }
    var arr = nums
    arr.sort()
    var sum = 0
    for i in 0..<arr.count {
        if i % 2 == 0 {
            sum += arr[i]
        }
    }
    return sum
}
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