zoukankan      html  css  js  c++  java
  • [Swift]LeetCode1122. 数组的相对排序 | Relative Sort Array

    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
    ➤微信公众号:山青咏芝(shanqingyongzhi)
    ➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/
    ➤GitHub地址:https://github.com/strengthen/LeetCode
    ➤原文地址:https://www.cnblogs.com/strengthen/p/11179476.html 
    ➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。
    ➤原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!
    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

    Given two arrays arr1 and arr2, the elements of arr2 are distinct, and all elements in arr2 are also in arr1.

    Sort the elements of arr1 such that the relative ordering of items in arr1 are the same as in arr2.  Elements that don't appear in arr2 should be placed at the end of arr1 in ascending order. 

    Example 1:

    Input: arr1 = [2,3,1,3,2,4,6,7,9,2,19], arr2 = [2,1,4,3,9,6]
    Output: [2,2,2,1,4,3,3,9,6,7,19] 

    Constraints:

    • arr1.length, arr2.length <= 1000
    • 0 <= arr1[i], arr2[i] <= 1000
    • Each arr2[i] is distinct.
    • Each arr2[i] is in arr1.

    给你两个数组,arr1 和 arr2

    • arr2 中的元素各不相同
    • arr2 中的每个元素都出现在 arr1 中

    对 arr1 中的元素进行排序,使 arr1 中项的相对顺序和 arr2 中的相对顺序相同。未在 arr2 中出现过的元素需要按照升序放在 arr1 的末尾。 

    示例:

    输入:arr1 = [2,3,1,3,2,4,6,7,9,2,19], arr2 = [2,1,4,3,9,6]
    输出:[2,2,2,1,4,3,3,9,6,7,19] 

    提示:

    • arr1.length, arr2.length <= 1000
    • 0 <= arr1[i], arr2[i] <= 1000
    • arr2 中的元素 arr2[i] 各不相同
    • arr2 中的每个元素 arr2[i] 都出现在 arr1 中

    8ms
     1 class Solution {
     2     func relativeSortArray(_ arr1: [Int], _ arr2: [Int]) -> [Int] {
     3       var dict = [Int: Int]()
     4       var result = [Int]()
     5       for a in arr1 {
     6         dict[a] = (dict[a] ?? 0)+1
     7       }
     8 
     9       for a in arr2 {
    10         if let val = dict[a] {
    11           for i in 0..<val {
    12             result.append(a)
    13           }
    14           dict.removeValue(forKey: a)
    15         }
    16       }
    17       var simpleArray = [Int]()
    18       if !dict.isEmpty {
    19         for (key,val) in dict {
    20           for i in 0..<val {
    21             simpleArray.append(key)
    22           }
    23           dict.removeValue(forKey: key)
    24         }
    25       }
    26       result.append(contentsOf: simpleArray.sorted())
    27       return result
    28     }
    29 }

    12ms

     1 class Solution {
     2     func relativeSortArray(_ arr1: [Int], _ arr2: [Int]) -> [Int] {
     3          var dict = Dictionary<Int,Int>()
     4         
     5         for value in arr1 {
     6             if dict[value] == nil {
     7                 dict[value] = 1
     8             }else {
     9                 dict[value] = dict[value]! + 1
    10             }
    11         }
    12         
    13         
    14         var outputArray: Array<Int> = Array<Int>()
    15         
    16         for value in arr2 {
    17             
    18             if let count = dict[value] {
    19                 for _ in 1...count{
    20                     outputArray.append(value)
    21                 }
    22             }
    23         }
    24         
    25         let set2 = Set(arr2)
    26         
    27        var appendArray = Array<Int>()
    28         
    29         for value in arr1 {
    30             if !set2.contains(value) {
    31                 appendArray.append(value)
    32             }
    33         }
    34         appendArray.sort()
    35         outputArray.append(contentsOf: appendArray)
    36         return outputArray        
    37     }
    38 }

    16ms

     1 class Solution {
     2     func relativeSortArray(_ arr1: [Int], _ arr2: [Int]) -> [Int] {
     3         let (numToCount, numsNotInArr2) = getNumToCount(arr1, arr2)
     4         var result = [Int]()
     5         for num in arr2 {
     6             if let count = numToCount[num] {
     7                 for _ in 0..<count {
     8                     result.append(num)
     9                 }
    10             }
    11         }
    12         result.append(contentsOf: numsNotInArr2)
    13         return result
    14     }
    15     
    16     private func getNumToCount(_ arr1: [Int], _ arr2: [Int]) -> ([Int:Int], [Int]) {
    17         let arr2Set = Set(arr2)
    18         var numToCount = [Int:Int]()
    19         var numsNotInArr2 = [Int]()
    20         for num in arr1 {
    21             if !arr2Set.contains(num) {
    22                 numsNotInArr2.append(num)
    23             } else {
    24                 numToCount.updateValue((numToCount[num] ?? 0) + 1, forKey: num)
    25             }
    26         }
    27         numsNotInArr2 = numsNotInArr2.sorted()
    28         return (numToCount, numsNotInArr2)
    29     }
    30 }

    Runtime: 20 ms

    Memory Usage: 21.2 MB
     1 class Solution {
     2     func relativeSortArray(_ arr1: [Int], _ arr2: [Int]) -> [Int] {
     3         var arr1 = arr1.sorted(by:<)
     4         var n:Int = arr1.count
     5         var used:[Bool] = [Bool](repeating:false,count:n)
     6         var result:[Int] = [Int]()
     7         for x in arr2
     8         {
     9             for i in 0..<n
    10             {
    11                 if arr1[i] == x && !used[i]
    12                 {
    13                     result.append(x)
    14                     used[i] = true
    15                 }
    16             }
    17         }
    18         for i in 0..<n
    19         {
    20             if !used[i]
    21             {
    22                 result.append(arr1[i])
    23             }
    24         }
    25         return result
    26     }
    27 }

    20ms
     1 class Solution {
     2   func relativeSortArray(_ arr1: [Int], _ arr2: [Int]) -> [Int] {
     3 
     4     var orderMap = [Int: Int]()
     5 
     6     for (offset, element) in arr2.enumerated() {
     7       orderMap[element] = offset
     8     }
     9 
    10     let result = arr1.sorted { left, right in
    11       switch (orderMap[left], orderMap[right]) {
    12       case (let leftOrder?, let rightOrder?):
    13         return leftOrder < rightOrder
    14       case (.none, .some):
    15         return false
    16       case (.some, .none):
    17         return true
    18       default:
    19         return left < right
    20       }
    21     }
    22     return result
    23   }
    24 }

    24ms

     1 class Solution {
     2     func relativeSortArray(_ arr1: [Int], _ arr2: [Int]) -> [Int] {
     3         var counter = [Int : Int]()
     4         for num in arr1 {
     5             counter[num, default: 0] += 1
     6         }
     7         
     8         var ans = [Int]()
     9         for num in arr2 {
    10             ans += Array(repeating: num, count: counter[num]!)
    11             counter[num] = nil
    12         }
    13         
    14         let nums = counter.keys.sorted()
    15         for num in nums {
    16             ans += Array(repeating: num, count: counter[num]!)
    17             counter[num] = nil
    18         }
    19         
    20         return ans
    21     }
    22 }

    28ms

     1 class Solution {
     2     func relativeSortArray(_ arr1: [Int], _ arr2: [Int]) -> [Int] {
     3         var dict = [Int: Int]()
     4         for i in 0..<arr2.count {
     5             dict[arr2[i]] = i
     6         }
     7         
     8         return arr1.sorted(by: { (e1, e2) in
     9             
    10             guard let idx1 = dict[e1] else {
    11                 if dict[e2] == nil { return e1 < e2 }
    12                 return false
    13             }
    14             guard let idx2 = dict[e2] else {
    15                 return true
    16             }
    17             return idx1 < idx2
    18         })       
    19         
    20     }
    21 }

    52ms

     1 class Solution {
     2   func relativeSortArray(_ arr1: [Int], _ arr2: [Int]) -> [Int] {
     3     var result = [Int]()
     4     var haystack = arr1
     5     for needle in arr2 {
     6       var index = 0
     7       for element in haystack {
     8         guard needle == element else { index += 1; continue }
     9         haystack.remove(at: index)
    10         result.append(element)
    11       }
    12     }
    13     result.append(contentsOf: haystack.sorted())
    14     return result
    15   }
    16 }
  • 相关阅读:
    小米手机4 rom 下载链接
    java 两个线程交替打印到100
    获取积分最多的人(二)
    刷题通过的题目排名
    考试分数(二)
    牛客的课程订单分析(二)
    牛客的课程订单分析(三)
    docker使用小记
    Dynamic Graph Collaborative Filtering
    Learnable Embedding Sizes for Recommender Systems
  • 原文地址:https://www.cnblogs.com/strengthen/p/11179476.html
Copyright © 2011-2022 走看看