[Swift]LeetCode721.账户合并|AccountsMerge

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Given a list accounts, each element accounts[i] is a list of strings, where the first element accounts[i][0] is a name, and the rest of the elements are emails representing emails of the account.

Now, we would like to merge these accounts. Two accounts definitely belong to the same person if there is some email that is common to both accounts. Note that even if two accounts have the same name, they may belong to different people as people could have the same name. A person can have any number of accounts initially, but all of their accounts definitely have the same name.

After merging the accounts, return the accounts in the following format: the first element of each account is the name, and the rest of the elements are emails in sorted order. The accounts themselves can be returned in any order.

Example 1:

Input: 
accounts = [["John", "[email protected]", "[email protected]"], ["John", "[email protected]"], ["John", "[email protected]", "[email protected]"], ["Mary", "[email protected]"]]
Output: [["John", '[email protected]', '[email protected]', '[email protected]'],  ["John", "[email protected]"], ["Mary", "[email protected]"]]
Explanation: 
The first and third John's are the same person as they have the common email "[email protected]".
The second John and Mary are different people as none of their email addresses are used by other accounts.
We could return these lists in any order, for example the answer [['Mary', '[email protected]'], ['John', '[email protected]'], 
['John', '[email protected]', '[email protected]', '[email protected]']] would still be accepted.

Note:

The length of accounts will be in the range [1, 1000].
The length of accounts[i] will be in the range [1, 10].
The length of accounts[i][j] will be in the range [1, 30].

给定一个列表 accounts，每个元素 accounts[i] 是一个字符串列表，其中第一个元素 accounts[i][0] 是 名称 (name)，其余元素是 emails 表示该帐户的邮箱地址。

现在，我们想合并这些帐户。如果两个帐户都有一些共同的邮件地址，则两个帐户必定属于同一个人。请注意，即使两个帐户具有相同的名称，它们也可能属于不同的人，因为人们可能具有相同的名称。一个人最初可以拥有任意数量的帐户，但其所有帐户都具有相同的名称。

合并帐户后，按以下格式返回帐户：每个帐户的第一个元素是名称，其余元素是按顺序排列的邮箱地址。accounts 本身可以以任意顺序返回。

例子 1:

Input: 
accounts = [["John", "[email protected]", "[email protected]"], ["John", "[email protected]"], ["John", "[email protected]", "[email protected]"], ["Mary", "[email protected]"]]
Output: [["John", '[email protected]', '[email protected]', '[email protected]'],  ["John", "[email protected]"], ["Mary", "[email protected]"]]
Explanation: 
  第一个和第三个 John 是同一个人，因为他们有共同的电子邮件 "[email protected]"。 
  第二个 John 和 Mary 是不同的人，因为他们的电子邮件地址没有被其他帐户使用。
  我们可以以任何顺序返回这些列表，例如答案[['Mary'，'[email protected]']，['John'，'[email protected]']，
  ['John'，'[email protected]'，'[email protected]'，'[email protected]']]仍然会被接受。

注意：

accounts的长度将在[1，1000]的范围内。
accounts[i]的长度将在[1，10]的范围内。
accounts[i][j]的长度将在[1，30]的范围内。

552ms

  1 class Solution {
  2     func accountsMerge(_ accounts: [[String]]) -> [[String]] {
  3         var dsu = DSU()
  4         for account in accounts {
  5             var emails = Array(account[1...])
  6             var name = account[0]
  7             dsu.addEdge(emails, name)
  8         }
  9         return dsu.unions()
 10     }
 11 }
 12 
 13 struct DSU {
 14     var mapping: [String: Int] = [:]
 15     var arr: [Int]
 16     var size: [Int]
 17     var comps: Int
 18     var names: [Int: String] = [:]
 19     
 20     mutating func unions() -> [[String]] {
 21         var res = [String: [String]]()
 22         for (email, index) in mapping {
 23             let parent = find(index)
 24             res["\(names[parent]!):\(parent)", default: []].append(email)
 25         }
 26         return res.map { arg0 in
 27             var (key, val) = arg0
 28             key = String(key.split(separator: ":")[0])
 29             return [key] + val.sorted()
 30         }
 31     }
 32     
 33     init() {
 34         arr = []
 35         size = []
 36         comps = 0
 37     }
 38     
 39     init(n: Int) {
 40         comps = n
 41         arr = Array(repeating: 0, count: n)
 42         size = Array(repeating: 1, count: n)
 43         for i in 0..<n {
 44             arr[i] = i
 45         }
 46     }
 47     
 48     mutating func addEdge(_ edge: [String], _ name: String) {
 49         if mapping[edge[0]] == nil {
 50             arr.append(arr.count)
 51             size.append(1)
 52             mapping[edge[0]] = arr.count - 1
 53             names[arr.count - 1] = name
 54         }
 55         for i in 1..<edge.count {
 56             if mapping[edge[i]] == nil {
 57                 arr.append(arr.count)
 58                 size.append(1)
 59                 mapping[edge[i]] = arr.count - 1
 60             }
 61             union(mapping[edge[0]]!, mapping[edge[i]]!)
 62         }
 63     }
 64     
 65     mutating func find(_ a: String, _ b: String) -> Bool {
 66         guard let indA = mapping[a], let indB = mapping[b] else { return false }
 67         return find(indA, indB)
 68     }
 69     
 70     mutating func find(_ a: Int) -> Int {
 71         return root(a)
 72     }
 73     
 74     mutating func union(_ a: Int, _ b: Int) {
 75         let rootA = root(a)
 76         let rootB = root(b)
 77         if rootA == rootB {
 78             return
 79         }
 80         if size[a] >= size[b] {
 81             arr[rootB] = rootA
 82             size[a] += size[b]
 83         } else {
 84             arr[rootA] = rootB
 85             size[b] += size[a]
 86         }
 87         comps -= 1
 88     }
 89     
 90     mutating func find(_ a: Int, _ b: Int) -> Bool {
 91         return root(a) == root(b)
 92     }
 93     
 94     private mutating func root(_ index: Int) -> Int {
 95         var index = index
 96         while arr[index] != index {
 97             arr[index] = arr[arr[index]]
 98             index = arr[index]
 99         }
100         return index
101     }
102 }

704ms

 1 class Solution {
 2     class UnionManager {
 3         var idMap = [String: Int]()
 4         var setMap = [Int]()
 5         
 6         func add(element: String) -> Int {
 7             if idMap[element] == nil{
 8                 let idForElement = setMap.count
 9                 idMap[element] = idForElement
10                 setMap.append(idForElement)
11             }
12             return idMap[element] ?? 0
13         }
14         
15         func findSet(id: Int) -> Int {
16             if setMap[id] != id {
17                 setMap[id] = findSet(id: setMap[id])
18             }
19             return setMap[id]
20         }
21         
22         func union(setOf string1: String, with string2: String) {
23             let id1 = add(element: string1)
24             let id2 = add(element: string2)
25             setMap[findSet(id: id1)] = findSet(id: id2)
26         }
27     }
28     
29     func accountsMerge(_ accounts: [[String]]) -> [[String]] {
30         let unionManager = UnionManager()
31         var emailToName = [String: String]()
32         let nonEmptyAccounts = accounts.filter{ $0.count > 1 }
33         for account in nonEmptyAccounts {
34             let name = account[0]
35             for email in account[1..<account.count] {
36                 emailToName[email] = name
37                 unionManager.union(setOf: email, with: account[1])
38             }
39         }
40         var results: [Int: [String]] = [:]
41         for (email, id) in unionManager.idMap {
42             let setId = unionManager.findSet(id: id)
43             results[setId, default: []].append(email)
44         }
45         return results.values.map { $0.sorted() }
46         .map { emailList in
47               if let name = emailToName[emailList[0]] {
48                   return [name] + emailList
49               } else {
50                   return emailList
51               }
52          }
53     }
54 }

732ms

  1 class Solution {
  2     func accountsMerge(_ accounts: [[String]]) -> [[String]] {
  3         guard accounts.count > 0 else {
  4             return []
  5         }
  6         let graph = Graph<String>(.undirected)
  7         var emailToNameDictionary: [String: String] = [:]
  8         for account in accounts {
  9             let name = account[0]
 10             let source = graph.createVertex(account[1])
 11             for i in 1..<account.count {
 12                 let email = account[i]
 13                 emailToNameDictionary[email] = name
 14                 if i > 1 {
 15                     let destination = graph.createVertex(account[i])
 16                     graph.addUndirectedEdge(between: source, and: destination)
 17                 }
 18             }
 19         }
 20         var mergedAccounts: [[String]] = []
 21         var discovered: Set<Vertex<String>> = []
 22         for vertex in graph.vertice {
 23             if !discovered.contains(vertex) {
 24                 var emails: [String] = []
 25                 depthFirstSearch(graph, vertex, &discovered, &emails)
 26                 let result = [emailToNameDictionary[vertex.val]!] + emails.sorted()
 27                 mergedAccounts.append(result)
 28             }
 29         }
 30         
 31         return mergedAccounts
 32     }
 33 }
 34 
 35 func depthFirstSearch<T: Hashable>(_ graph: Graph<T>, 
 36                                    _ source: Vertex<T>, 
 37                                    _ discovered: inout Set<Vertex<T>>, 
 38                                    _ emails: inout [T]
 39                                   ) {
 40     discovered.insert(source)
 41     emails.append(source.val)
 42     for edge in graph.edges(of: source) {
 43         let destination = edge.destination
 44         if !discovered.contains(destination) {
 45             depthFirstSearch(graph, destination, &discovered, &emails)
 46         }
 47     }
 48 }
 49 
 50 
 51 enum GraphType {
 52     case directed 
 53     case undirected 
 54 }
 55 
 56 struct Vertex<T: Hashable>: Hashable {
 57     public var val: T 
 58     public init(_ val: T) {
 59         self.val = val
 60     }
 61     
 62     public var hashValue : Int {
 63         return val.hashValue
 64     }
 65     
 66     public static func ==(lhs: Vertex<T>, rhs: Vertex<T>) -> Bool {
 67         return lhs.val == rhs.val
 68     }
 69 }
 70 
 71 struct Edge<T: Hashable> {
 72     public let source: Vertex<T>
 73     public let destination: Vertex<T>
 74     public let weight: Double? 
 75     public init(source: Vertex<T>, destination: Vertex<T>, weight: Double? = nil ) {
 76         self.source = source 
 77         self.destination = destination 
 78         self.weight = weight 
 79     }
 80 }
 81 
 82 
 83 class Graph<T: Hashable> {
 84     public var vertice: [Vertex<T>] = []
 85     private var adjacencyList: [Vertex<T>: [Edge<T>]] = [:]
 86     public var type: GraphType 
 87     
 88     public init(_ type: GraphType) {
 89         self.type = type
 90     }
 91     
 92     public func createVertex(_ val: T) -> Vertex<T> {
 93         let source = Vertex(val)
 94         if adjacencyList[source] == nil {
 95             adjacencyList[source] = []
 96             vertice.append(source)
 97         }
 98         return source 
 99     }
100     
101     public func addDirectedEdge(from source: Vertex<T>, 
102                                 to destination: Vertex<T>, 
103                                 weight: Double? = nil 
104                                ) {
105         let edge = Edge(source: source, destination: destination, weight: weight)
106         adjacencyList[source]?.append(edge)
107     }
108     
109     public func addUndirectedEdge(between source: Vertex<T>,
110                                   and destination: Vertex<T>, 
111                                   weight: Double? = nil 
112                                  ) {
113         addDirectedEdge(from: source, to: destination, weight: weight)
114         addDirectedEdge(from: destination, to: source, weight: weight)
115     }
116     
117     public func edges(of source: Vertex<T>) -> [Edge<T>] {
118         return adjacencyList[source] ?? []
119     }
120     
121     public func weight(from source: Vertex<T>, 
122                        to destination: Vertex<T>
123                       ) -> Double? {
124         return adjacencyList[source]?.first{ $0.destination == destination }?.weight 
125     }
126 }

852ms

 1 class Solution {
 2 // father[儿子] = 老大哥
 3 var father = [Int: Int]()
 4 
 5 func accountsMerge(_ accounts: [[String]]) -> [[String]] {
 6     
 7     // 1 union
 8     let emailToID = getEmailToID(accounts)
 9     for email in emailToID.keys {
10         let ids = emailToID[email]!
11         for i in ids {
12             union(i, ids[0])
13         }
14     }
15     
16     //2 merge
17     let idToEmailSet = getIdToEmailSet(accounts)
18     var mergedAccounts = [[String]]()
19     
20     for id in idToEmailSet.keys {
21         var sortedEmails = idToEmailSet[id]!.sorted()
22         sortedEmails.insert(accounts[id][0], at: 0)
23         mergedAccounts.append(sortedEmails)
24     }
25     
26     return mergedAccounts
27 }
28 
29 //find -> 找老大哥
30 func find(_ id: Int) -> Int {
31     var id = id
32     
33     var path = [Int]()
34     while let nextID = father[id] {
35         path.append(id)
36         id = nextID
37         
38     }
39     
40     for i in path {
41         father[i] = id
42     }
43     
44     return id
45 }
46 
47 // union
48 func union(_ a: Int, _ b: Int) {
49     let rootA = find(a)
50     let rootB = find(b)
51     if rootA != rootB {
52         father[rootA] = rootB
53     }
54 }
55 
56 
57 // 这个只是data处理
58 func getEmailToID(_ accounts: [[String]]) -> [String: [Int]] {
59     var emailToID = [String: [Int]]()
60     
61     for (userID, acnt) in accounts.enumerated() {
62         for i in 1..<acnt.count {
63             let curEmail = acnt[i]
64             var ids = emailToID[curEmail, default: [Int]()]
65             ids.append(userID)
66             emailToID[curEmail] = ids
67         }
68     }
69     
70     return emailToID
71 }
72 
73 func getIdToEmailSet(_ accounts: [[String]]) -> [Int: Set<String>] {
74     var idToEmailSet = [Int: Set<String>]()
75     for id in 0..<accounts.count {
76         let root_id = find(id)
77         var emailSet = idToEmailSet[root_id, default: Set<String>()]
78         
79         let account = accounts[id]
80         for i in 1..<account.count {
81             emailSet.insert(account[i])
82         }
83         idToEmailSet[root_id] = emailSet
84     }
85     
86     return idToEmailSet
87   }
88 }

864ms

 1 class Solution {
 2     func accountsMerge(_ accounts: [[String]]) -> [[String]] {
 3         guard accounts.count > 0 else {
 4             return []
 5         }
 6         
 7         var emailToName: [String: String] = [:]
 8         var unionFind = UnionFind<String>()
 9         for account in accounts {
10             let name = account[0]
11             let firstEmail = account[1]
12             unionFind.addSet(firstEmail)
13             for i in 1..<account.count {
14                 let email = account[i]
15                 emailToName[email] = name
16                 
17                 if i > 1 {
18                     let email = account[i]
19                     unionFind.addSet(email)
20                     unionFind.union(firstEmail, email)
21                 }
22             }
23         }
24         
25         var mergedAccount: [[String]] = []
26         var storage: [Int: [String]] = [:]
27         for element in unionFind.index.keys {
28             let parentIndex = unionFind.find(element)!
29             if storage[parentIndex] == nil {
30                 storage[parentIndex] = [element]
31             } else {
32                 storage[parentIndex]!.append(element)
33             }
34         }
35         
36         for emails in storage.values {
37             let name = emailToName[emails.first!]!
38