(未完成)332. Reconstruct Itinerary

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.

s思路:
1. 给两两连接的机场，要求得到一整条路线。仔细观察，还是图的遍历嘛，但又不同与之前看到的图的遍历，因为之前的遍历是考虑节点是否被访问，而这里则是边是否被访问。这一点一定要先搞明白。如下图，从JFK开始，有两个邻居，ATL和SFO,先访问哪个题目有规定，字典排序小的机场先访问，所以先访问ATL,由于是找路线，自然是用dfs最方便，访问了ATL，常规的做法是需要标记已经访问过，但这里只需要使用一次这条边，所以干脆直接删除这条边，就不用标记是否访问，然后到ATL继续遍历，发现也有两个邻居JFK,SFO，这里也是一样，西安访问JFK,同时把JFK这条边删除；现在到JFK后，发现JFK只有一个邻居SFO（ATL已经被删除），于是访问SFO,同时删除SFO这条链接；继续遍历SFO的邻居，发现有ATL,于是访问ATL,同时删除ATL这条边，现在到ATL发现也只有一个邻居SFO,于是继续访问SFO,同时删除链接，到SFO发现没有邻居了，意味着结束了吗？还是需要回到上一层继续遍历呢？保险起见，是需要回到上一层吧！

2.由于要字典顺序访问，所以最好对每个节点的邻居排序，所以不用vector放邻居，而用map.
3. 上面分析还是不对，debug的时候，发现前面分析的不对呀，没有反应过来，所谓图里面元素的排序，本质就是topological sort,先用dfs走一边，谁的邻居节点都遍历完谁就先放入vector，最好等所有节点都放入vector再reverse vector即可！由于不是检测是否有环，所以仍然不需要用visited来表示一个节点是否访问完全还是正在访问，

class Solution {public:    void dfs(unordered_map<string,set<string>>&mm,vector<string>&res,string cur){        if(mm.count(cur)==0) return;            for(auto it=mm[cur].begin();it!=mm[cur].end();)        for(auto&str:mm[cur]){            string next=str;            mm[cur].erase(str);            res.push_back(next);            dfs(mm,res,next);        }    }       vector<string> findItinerary(vector<pair<string, string>> tickets) {        //        unordered_map<string,set<string>> mm;        for(auto&t:tickets){            mm[t.first].insert(t.second);                }        vector<string> res;        dfs(mm,res,"JFK");        return res;    }};