Question

332. 重新安排行程

English Version

题目描述

给你一份航线列表 tickets ，其中 tickets[i] = [fromi, toi] 表示飞机出发和降落的机场地点。请你对该行程进行重新规划排序。

所有这些机票都属于一个从 JFK（肯尼迪国际机场）出发的先生，所以该行程必须从 JFK 开始。如果存在多种有效的行程，请你按字典排序返回最小的行程组合。

• 例如，行程 ["JFK", "LGA"] 与 ["JFK", "LGB"] 相比就更小，排序更靠前。

假定所有机票至少存在一种合理的行程。且所有的机票 必须都用一次 且 只能用一次。

 

示例 1：

输入：tickets = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
输出：["JFK","MUC","LHR","SFO","SJC"]

示例 2：

输入：tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
输出：["JFK","ATL","JFK","SFO","ATL","SFO"]
解释：另一种有效的行程是 ["JFK","SFO","ATL","JFK","ATL","SFO"] ，但是它字典排序更大更靠后。

 

提示：

• 1 <= tickets.length <= 300
• tickets[i].length == 2
• fromi.length == 3
• toi.length == 3
• fromi 和 toi 由大写英文字母组成
• fromi != toi

解法

方法一

class Solution:
  def findItinerary(self, tickets: List[List[str]]) -> List[str]:
    graph = defaultdict(list)

    for src, dst in sorted(tickets, reverse=True):
      graph[src].append(dst)

    itinerary = []

    def dfs(airport):
      while graph[airport]:
        dfs(graph[airport].pop())
      itinerary.append(airport)

    dfs("JFK")

    return itinerary[::-1]

class Solution {
  void dfs(Map<String, Queue<String>> adjLists, List<String> ans, String curr) {
    Queue<String> neighbors = adjLists.get(curr);
    if (neighbors == null) {
      ans.add(curr);
      return;
    }
    while (!neighbors.isEmpty()) {
      String neighbor = neighbors.poll();
      dfs(adjLists, ans, neighbor);
    }
    ans.add(curr);
    return;
  }

  public List<String> findItinerary(List<List<String>> tickets) {
    Map<String, Queue<String>> adjLists = new HashMap<>();
    for (List<String> ticket : tickets) {
      String from = ticket.get(0);
      String to = ticket.get(1);
      if (!adjLists.containsKey(from)) {
        adjLists.put(from, new PriorityQueue<>());
      }
      adjLists.get(from).add(to);
    }
    List<String> ans = new ArrayList<>();
    dfs(adjLists, ans, "JFK");
    Collections.reverse(ans);
    return ans;
  }
}

class Solution {
public:
  vector<string> findItinerary(vector<vector<string>>& tickets) {
    unordered_map<string, priority_queue<string, vector<string>, greater<string>>> g;
    vector<string> ret;

    // Initialize the graph
    for (const auto& t : tickets) {
      g[t[0]].push(t[1]);
    }

    findItineraryInner(g, ret, "JFK");

    ret = {ret.rbegin(), ret.rend()};

    return ret;
  }

  void findItineraryInner(unordered_map<string, priority_queue<string, vector<string>, greater<string>>>& g, vector<string>& ret, string cur) {
    if (g.count(cur) == 0) {
      // This is the end point
      ret.push_back(cur);
      return;
    } else {
      while (!g[cur].empty()) {
        auto front = g[cur].top();
        g[cur].pop();
        findItineraryInner(g, ret, front);
      }
      ret.push_back(cur);
    }
  }
};