332. Reconstruct Itinerary

greatwhole 發布于2019-08-14 14:17 / 1301人閱讀

摘要：題目解答都是用來解，一個用一個用來實現深度優先搜索，搜索到一個城市只是的時候即沒有出度的時候，把這個記入中去，因為它肯定是最后到達的城市，然后依次向前類推的要求在存入的時候就用先存好先進去的說明是出發城市，那么最先出發的城市最后出來

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> graph, List result) {
    //深度優先搜索，搜索到一個城市只是arrival city的時候（即沒有出度的時候，把這個city記入list中去，因為它肯定是最后到達的城市，然后依次向前類推
    PriorityQueue arrivals = graph.get(departure);
    while (arrivals != null && !arrivals.isEmpty()) {
        dfs(arrivals.poll(), graph, result);
    }
    result.add(0, departure);
}

public List findItinerary(String[][] tickets) {
    List result = new ArrayList();
    //lexical order的要求在存入graph的時候就用priority queue先存好
    Map> graph = new HashMap<>();
    for (String[] iter : tickets) {
        graph.putIfAbsent(iter[0], new PriorityQueue());
        graph.get(iter[0]).add(iter[1]);
    }
    
    dfs("JFK", graph, result);
    return result;
}

Stack version:

public List findItinerary(String[][] tickets) {
    List result = new ArrayList();
    Map> graph = new HashMap();
    for (String[] iter : tickets) {
        graph.putIfAbsent(iter[0], new PriorityQueue());
        graph.get(iter[0]).add(iter[1]);
    }
    
    Stack stack = new Stack();
    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，那么最先出發的城市最后出來
            stack.push(graph.get(stack.peek()).poll());
        }
        result.add(0, stack.pop());
    }
    return result;
}

GPU云服務器云服務器 Itinerary Reconstruct

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