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"] has a smaller lexical order than ["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: tickets = [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]] Return ["JFK", "MUC", "LHR", "SFO", "SJC"].
Example 2: tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]] Return ["JFK","ATL","JFK","SFO","ATL","SFO"]. Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"]. But it is larger in lexical order.
Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.
public class Solution{ List<String> res = new ArrayList<String>(); Map<String, PriorityQueue<String>> map = new HashMap<String, PriorityQueue<String>>(); String[][] ts; public List<String> findItinerary(String[][] tickets) { ts = tickets; if(ts==null || ts.length==0) return res; int m = ts.length; for(int i=0; i<m; i++){ String from = ts[i][0]; String to = ts[i][1]; if(!map.containsKey(from)){ map.put(from, new PriorityQueue<String>()); } map.get(from).add(to); } DFS("JFK"); return res; } public void DFS(String cur){ while(map.containsKey(cur) && !map.get(cur).isEmpty()){ //因为构造的是一个解，不需要记录不同DFS的路径 DFS(map.get(cur).poll()); } res.add(0, cur); } } public class Solution{ LinkedList<String> res = new LinkedList<String>(); Map<String, PriorityQueue<String>> map = new HashMap<String, PriorityQueue<String>>(); String[][] ts; public List<String> findItinerary(String[][] tickets) { ts = tickets; int m = ts.length; if(ts==null || ts.length==0) return new LinkedList<String>(); for(String[] t:ts){ if(!map.containsKey(t[0])){ map.put(t[0], new PriorityQueue<String>()); } map.get(t[0]).offer(t[1]); } DFS("JFK"); return res; } public void DFS(String cur){ while(map.containsKey(cur)&& !map.get(cur).isEmpty()){ DFS(map.get(cur).poll()); } res.addFirst(cur); } }
