332. Reconstruct Itinerary 难度：medium 类别：图

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luoyingmin

最新推荐文章于 2025-04-27 08:12:49 发布

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本文介绍了一种解决给定一系列航空机票后如何重构行程路径的问题。通过使用栈与哈希映射，结合字典序最小原则，确保从起点JFK出发能够获得一条有效的旅行路线。

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题目：

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.

思路：

题目要求我们根据所给的机票信息找到一条从起点到终点的路径，因为题目保证路径肯定存在，所以从JFK开始，每次选取最小路径走，如果走不下去，只有可能遇到了终结点，这样就形成了从JFK到PP的主路径。剩下没走的边只有可能形成环，只要将环并入到主路径上就完成了最终路径的搜索。

程序：

class Solution {
public:
    vector<string> findItinerary(vector<pair<string, string>> tickets) {
        unordered_map<string, multiset<string>> m;
        vector<string> res;
        if (tickets.size() <= 0) {
            return res;
        }
        for (pair<string, string> p: tickets) {
            m[p.first].insert(p.second);

        }
        stack<string> s;
        s.push("JFK");
        while (s.size()) {
            string next = s.top();
            if (m[next].empty()) {
                res.push_back(next);
                s.pop();
            } else {
                s.push(*m[next].begin());
                m[next].erase(m[next].begin());
            }
        }
        reverse(res.begin(), res.end());
        return res;
    }
};