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.

题意分析

题目意思是说，给定一系列的飞机出发地和飞机目的地，求从JFK出发的行程，这个行程的结果要求是最小序列（如果有多个可能行程序列的话）

思路分析

这道题跟course schedule的思路分析相似，course schedule的解法是针对点的dfs，这道题解法是针对边的dfs，在这里提供另外一种类似的思路，利用栈来进行回溯，因为题目要求是最小序列，因此可以使用multiset来存储这个图。开始遍历图中的边，只要图中的边没有遍历完，那么就把遍历到的点加进栈中，然后继续遍历，最后把栈的点依次加入到vector的首位，则得到最终结果

AC代码

class Solution {
public:
    vector<string> findItinerary(vector<pair<string, string>> tickets) {
        vector<string> result;
        stack<string> backtrack;
        backtrack.push("JFK");
        unordered_map<string, multiset<string>> mapstr;
        for (int i = 0; i < tickets.size(); ++i) {
            mapstr[tickets[i].first].insert(tickets[i].second);
        }
        while (!backtrack.empty()) {
            string currentdes = backtrack.top();
            if (mapstr[currentdes].empty()) {
                result.insert(result.begin(), currentdes);
                backtrack.pop();
            } else {
                backtrack.push(*mapstr[currentdes].begin());
                mapstr[currentdes].erase(mapstr[currentdes].begin());
            }
        }
        return result;
    }
};