332. Reconstruct Itinerary ----------- H ！！

You are given a list of airline tickets where tickets[i] = [fromi, toi] represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.

All of the tickets belong to a man who departs from "JFK", thus, the itinerary must begin with "JFK". 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"].

You may assume all tickets form at least one valid itinerary. You must use all the tickets once and only once.

 

Example 1:

Input: tickets = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
Output: ["JFK","MUC","LHR","SFO","SJC"]

Example 2:

Input: tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
Output: ["JFK","ATL","JFK","SFO","ATL","SFO"]
Explanation: Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"] but it is larger in lexical order.

 

Constraints:

• 1 <= tickets.length <= 300
• tickets[i].length == 2
• fromi.length == 3
• toi.length == 3
• fromi and toi consist of uppercase English letters.
• fromi != toi

A:

没啥说的， 就是经典的DFS。 比较tricky的地方，就是需要trial，if failed, need reset

class Solution {
public:
    vector<string> findItinerary(vector<vector<string>>& tickets) {
        unordered_map<string, vector<string>> M;
        int N = tickets.size();
        for (auto ticket : tickets) {
            M[ticket[0]].push_back(ticket[1]);
        }
        for (auto& p : M) {
            sort(p.second.rbegin(), p.second.rend()); // reverse order
        }
        vector<string> res{"JFK"};
        dfs(M, res, N + 1);
        return res;
    }
private:
    void dfs(unordered_map<string, vector<string>>& M, vector<string>& res,
             int N) {
        string from = res.back();
        auto& toCities = M[from];
        for (int i = toCities.size()-1; i>=0 ; i--) {
            res.push_back(toCities[i]);
            toCities.erase(toCities.begin() + i);
            dfs(M, res, N);
            if (res.size() == N)
                return;
            else { // need trial and error)
                toCities.insert(toCities.begin() + i, res.back());
                res.pop_back();
            }
        }
    }
};

然而， 上面的dfs。LTE 。坏处就是需要删除vector的某个节点

那么，我们不删除， 而只是mark呢？

class Solution {
public:
    vector<string> findItinerary(vector<vector<string>>& tickets) {
        unordered_map<string, vector<string>> M;
        int N = tickets.size();
        for (auto ticket : tickets) {
            M[ticket[0]].push_back(ticket[1]);
        }
        for (auto& p : M) {
            sort(p.second.rbegin(), p.second.rend()); // reverse order
        }
        vector<string> res{"JFK"};
        dfs(M, res, N + 1);
        return res;
    }
private:
    void dfs(unordered_map<string, vector<string>>& M, vector<string>& res,
             int N) {
        string from = res.back();
        auto& toCities = M[from];
        for (int i = toCities.size() - 1; i >= 0; i--) {
            if(toCities[i]=="") // instead of erase, we mark it
                continue;
            res.push_back(toCities[i]);
            toCities[i] = "";
            dfs(M, res, N);
            if (res.size() == N)
                return;
            else { // need trial and error)
                toCities[i] = res.back();
                res.pop_back();
            }
        }
    }
};

依然LTE。 

看了fail的例子。加了2行

            if(i+1 < toCities.size() && toCities[i] == toCities[i+1]) // skil same choice
                continue;

最终的pass的代码是： 
class Solution {
public:
    vector<string> findItinerary(vector<vector<string>>& tickets) {
        unordered_map<string, vector<string>> M;
        int N = tickets.size();
        for (auto ticket : tickets) {
            M[ticket[0]].push_back(ticket[1]);
        }
        for (auto& p : M) {
            sort(p.second.rbegin(), p.second.rend()); // reverse order
        }
        vector<string> res{"JFK"};
        dfs(M, res, N + 1);
        return res;
    }
private:
    void dfs(unordered_map<string, vector<string>>& M, vector<string>& res,
             int N) {
        string from = res.back();
        auto& toCities = M[from];
        for (int i = toCities.size() - 1; i >= 0; i--) {
            if(toCities[i]=="") // instead of erase, we mark it
                continue;
            if(i+1 < toCities.size() && toCities[i] == toCities[i+1]) // same choice
                continue;
            res.push_back(toCities[i]);
            toCities[i] = "";
            dfs(M, res, N);
            if (res.size() == N)
                return;
            else { // need trial and error)
                toCities[i] = res.back();
                res.pop_back();
            }
        }
    }
};

Mistakes:

Inside your DFS loop, you're doing this:

auto V = M[res.back()];

This creates a copy of the vector, so changes like:

V.erase(...);

are only applied to the copy, not the actual M[...].

As a result:

• You're never really updating the flight map.

• Backtracking is not working as intended.

---

✅ Fix:

You need to access and modify the original vector by reference:

auto& V = M[res.back()];