332. Reconstruct Itinerary(重要)

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mb63887cf57331d 2022-12-02 00:48:19 ©著作权

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[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.

class Solution {
public:
  vector<string> findItinerary(vector<pair<string, string>> tickets) {
    map<string, multiset<string>> edges;
    for (auto p : tickets){
      edges[p.first].insert(p.second);
    }
    dfs(edges, "JFK");
    return vector<string>(res.rbegin(),res.rend());
  }
private:
  void dfs(map<string, multiset<string>> &edges, string str){
    while (edges[str].size()) {
      string t = *edges[str].begin();
      edges[str].erase(edges[str].begin());
      dfs(edges, t);
    }
    res.push_back(str);//放在后面
  }
  vector<string> res;
};