[Swift]LeetCode332. 重新安排行程 | Reconstruct Itinerary
阿新 • • 發佈:2019-01-13
Given a list of airline tickets represented by pairs of departure and arrival airports [from, to]
, reconstruct the itinerary in order. All of the tickets belong to a man who departs from JFK
. Thus, the itinerary must begin with JFK
.
Note:
- If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string. For example, the itinerary
["JFK", "LGA"]
["JFK", "LGB"]
. - All airports are represented by three capital letters (IATA code).
- You may assume all tickets form at least one valid itinerary.
Example 1:
Input:
[["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Output:["JFK", "MUC", "LHR", "SFO", "SJC"]
Example 2:
Input:
[["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
Output:["JFK","ATL","JFK","SFO","ATL","SFO"]
Explanation: Another possible reconstruction is["JFK","SFO","ATL","JFK","ATL","SFO"]
. But it is larger in lexical order.
給定一個機票的字串二維陣列 [from, to]
說明:
- 如果存在多種有效的行程,你可以按字元自然排序返回最小的行程組合。例如,行程 ["JFK", "LGA"] 與 ["JFK", "LGB"] 相比就更小,排序更靠前
- 所有的機場都用三個大寫字母表示(機場程式碼)。
- 假定所有機票至少存在一種合理的行程。
示例 1:
輸入:
[["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
輸出:["JFK", "MUC", "LHR", "SFO", "SJC"]
示例 2:
輸入:
[["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
輸出:["JFK","ATL","JFK","SFO","ATL","SFO"]
解釋: 另一種有效的行程是["JFK","SFO","ATL","JFK","ATL","SFO"]
。但是它自然排序更大更靠後。
144ms
1 class Solution { 2 var graph: [String: [String]] = [:] 3 var route: [String] = [] 4 func findItinerary(_ tickets: [[String]]) -> [String] { 5 6 for t in tickets { 7 var city = graph[t[0]] ?? [] 8 city.append(t[1]) 9 graph[t[0]] = city 10 } 11 12 for key in graph.keys { 13 graph[key]?.sort() 14 } 15 self.visit("JFK") 16 return self.route 17 } 18 19 private func visit(_ city: String) { 20 if let dests = graph[city] { 21 while !graph[city]!.isEmpty { 22 let dest = graph[city]!.removeFirst() 23 self.visit(dest) 24 } 25 } 26 self.route.insert(city, at: 0) 27 } 28 }
568ms
1 class Solution { 2 func findItinerary(_ tickets: [[String]]) -> [String] { 3 4 let graph = buildGraph(tickets) 5 var res = [String]() 6 dfs(graph, from: "JFK",0, &res) 7 8 return res 9 } 10 11 @discardableResult func dfs(_ graph : Graph<String>, from : String ,_ visitedCount : Int, _ res : inout [String]) -> Bool { 12 res.append(from) 13 if visitedCount == graph.edges.count { 14 return true 15 } 16 let edges = graph.edgesFrom(graph.createVertex(from)).sorted{$0.to.data < $1.to.data} 17 for edge in edges where edge.visited == false { 18 edge.visited = true 19 if dfs(graph, from: edge.to.data, visitedCount+1, &res) == true { 20 return true 21 } 22 edge.visited = false 23 } 24 res.removeLast() 25 return false 26 27 } 28 29 func buildGraph(_ tickets : [[String]]) -> Graph<String> { 30 let graph = Graph<String>() 31 for i in 0..<tickets.count { 32 let ticket = tickets[i] 33 let from = graph.createVertex(ticket[0]) 34 let to = graph.createVertex(ticket[1]) 35 graph.addDirectedEdge(from, to: to, withWeight: Double(i)) 36 } 37 return graph 38 } 39 //===================實現Graph的分割線====================== 40 struct Vertex<T> : Equatable,Hashable where T : Hashable { 41 var data : T 42 var index : Int 43 44 public var description: String { 45 return "\(index):\(data)" 46 } 47 } 48 49 public class Edge<T>: Equatable,Hashable where T: Hashable { 50 public static func == (lhs: Solution.Edge<T>, rhs: Solution.Edge<T>) -> Bool { 51 return lhs.from == rhs.from && lhs.to == rhs.to && lhs.weight == rhs.weight 52 } 53 54 public let from: Vertex<T> 55 public let to: Vertex<T> 56 57 public let weight:Double? 58 59 public var visited : Bool 60 61 public var hashValue:Int { 62 var string = "\(from.description)\(to.description)" 63 if weight != nil{ 64 string.append("\(weight!)") 65 } 66 return string.hashValue 67 } 68 69 init(from: Vertex<T>, to: Vertex<T>, weight: Double?, visited: Bool) { 70 self.from = from 71 self.to = to 72 self.weight = weight 73 self.visited = visited 74 } 75 } 76 77 78 79 private class EdgeList<T> where T: Hashable { 80 81 var vertex: Vertex<T> 82 var edges: [Edge<T>]? 83 84 init(vertex: Vertex<T>) { 85 self.vertex = vertex 86 } 87 88 func addEdge(_ edge: Edge<T>) { 89 edges?.append(edge) 90 } 91 92 } 93 94 95 class Graph<T> where T : Hashable { 96 var vertices : [Vertex<T>] { 97 var vertices = [Vertex<T>]() 98 for edgeList in adjacencyList { 99 vertices.append(edgeList.vertex) 100 } 101 return vertices 102 } 103 104 private var adjacencyList: [EdgeList<T>] = [] 105 106 open var edges: [Edge<T>] = [Edge<T>]() 107 108 open func addDirectedEdge(_ from: Vertex<T>, to: Vertex<T>, withWeight weight: Double?) { 109 // works 110 let edge = Edge(from: from, to: to, weight: weight, visited: false) 111 let edgeList = adjacencyList[from.index] 112 if edgeList.edges != nil { 113 edgeList.addEdge(edge) 114 } else { 115 edgeList.edges = [edge] 116 } 117 edges.append(edge) 118 } 119 120 open func createVertex(_ data: T) -> Vertex<T> { 121 122 let matchingVertices = vertices.filter { vertex in 123 return vertex.data == data 124 } 125 126 if matchingVertices.count > 0 { 127 return matchingVertices.last! 128 } 129 130 let vertex = Vertex(data: data, index: adjacencyList.count) 131 adjacencyList.append(EdgeList(vertex: vertex)) 132 return vertex 133 } 134 135 open func edgesFrom(_ sourceVertex: Vertex<T>) -> [Edge<T>] { 136 return adjacencyList[sourceVertex.index].edges ?? [] 137 } 138 139 } 140 }