332. Reconstruct Itinerary

NO IMAGE

題目:
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.

解答:
都是用DFS來解,一個用recursion, 一個用stack來實現:
Recursion version:

public void dfs(String departure, Map<String, PriorityQueue<String>> graph, List<String> result) {
//深度優先搜尋,搜尋到一個城市只是arrival city的時候(即沒有出度的時候,把這個city記入list中去,因為它肯定是最後到達的城市,然後依次向前類推
PriorityQueue<String> arrivals = graph.get(departure);
while (arrivals != null && !arrivals.isEmpty()) {
dfs(arrivals.poll(), graph, result);
}
result.add(0, departure);
}
public List<String> findItinerary(String[][] tickets) {
List<String> result = new ArrayList<String>();
//lexical order的要求在存入graph的時候就用priority queue先存好
Map<String, PriorityQueue<String>> graph = new HashMap<>();
for (String[] iter : tickets) {
graph.putIfAbsent(iter[0], new PriorityQueue<String>());
graph.get(iter[0]).add(iter[1]);
}
dfs("JFK", graph, result);
return result;
}

Stack version:

public List<String> findItinerary(String[][] tickets) {
List<String> result = new ArrayList<String>();
Map<String, PriorityQueue<String>> graph = new HashMap();
for (String[] iter : tickets) {
graph.putIfAbsent(iter[0], new PriorityQueue<String>());
graph.get(iter[0]).add(iter[1]);
}
Stack<String> stack = new Stack<String>();
stack.push("JFK");
while (!stack.isEmpty()) {
while (graph.containsKey(stack.peek()) && !graph.get(stack.peek()).isEmpty()) {
//先進去的說明是出發城市,那麼最先出發的城市最後出來
stack.push(graph.get(stack.peek()).poll());
}
result.add(0, stack.pop());
}
return result;
}