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  • [Swift]LeetCode1213. 三个有序数组的交集 | Intersection of Three Sorted Arrays

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    Given three integer arrays arr1arr2 and arr3 sorted in strictly increasing order, return a sorted array of only the integers that appeared in all three arrays.

    Example 1:

    Input: arr1 = [1,2,3,4,5], arr2 = [1,2,5,7,9], arr3 = [1,3,4,5,8]
    Output: [1,5]
    Explanation: Only 1 and 5 appeared in the three arrays. 

    Constraints:

    • 1 <= arr1.length, arr2.length, arr3.length <= 1000
    • 1 <= arr1[i], arr2[i], arr3[i] <= 2000

    给出三个均为 严格递增排列 的整数数组 arr1arr2 和 arr3

    返回一个由 仅 在这三个数组中 同时出现 的整数所构成的有序数组。 

    示例:

    输入: arr1 = [1,2,3,4,5], arr2 = [1,2,5,7,9], arr3 = [1,3,4,5,8]
    输出: [1,5]
    解释: 只有 1 和 5 同时在这三个数组中出现.

    提示:

    1. 1 <= arr1.length, arr2.length, arr3.length <= 1000
    2. 1 <= arr1[i], arr2[i], arr3[i] <= 2000

    Runtime: 84 ms
    Memory Usage: 21.5 MB
     1 class Solution {
     2     func arraysIntersection(_ arr1: [Int], _ arr2: [Int], _ arr3: [Int]) -> [Int] {
     3         var m:[Int:Int] = [Int:Int]()
     4         for a in arr1 {m[a,default:0] += 1}
     5         for a in arr2 {m[a,default:0] += 1}
     6         for a in arr3 {m[a,default:0] += 1}
     7         var res:[Int] = [Int]()
     8         for (key,val) in m
     9         {
    10             if val == 3
    11             {
    12                 res.append(key)
    13             }
    14         }
    15         return res.sorted(by:<)
    16     }
    17 }

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  • 原文地址:https://www.cnblogs.com/strengthen/p/11608027.html
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