[LeetCode] Reconstruct Itinerary 重建行程單

 

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:

1. 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"].
2. All airports are represented by three capital letters (IATA code).
3. You may assume all tickets may 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.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

 

這道題給咱們一堆飛機票，讓咱們創建一個行程單，若是有多種方法，取其中字母順序小的那種方法。這道題的本質是有向圖的遍歷問題，那麼LeetCode關於有向圖的題只有兩道Course Schedule和Course Schedule II，而那兩道是關於有向圖的頂點的遍歷的，而本題是關於有向圖的邊的遍歷。每張機票都是有向圖的一條邊，咱們須要找出一條通過全部邊的路徑，那麼DFS不是咱們的不二選擇。先來看遞歸的結果，咱們首先把圖創建起來，經過鄰接鏈表來創建。因爲題目要求解法按字母順序小的，那麼咱們考慮用multiset，能夠自動排序。等咱們圖創建好了之後，從節點JFK開始遍歷，只要當前節點映射的multiset裏有節點，咱們取出這個節點，將其在multiset裏刪掉，而後繼續遞歸遍歷這個節點，因爲題目中限定了必定會有解，那麼等圖中全部的multiset中都沒有節點的時候，咱們把當前節點存入結果中，而後再一層層回溯回去，將當前節點都存入結果，那麼最後咱們結果中存的順序和咱們須要的相反的，咱們最後再翻轉一下便可，參見代碼以下：

 

解法一：

class Solution {
public:
    vector<string> findItinerary(vector<pair<string, string>> tickets) {
        vector<string> res;
        unordered_map<string, multiset<string>> m;
        for (auto a : tickets) {
            m[a.first].insert(a.second);
        }
        dfs(m, "JFK", res);
        return vector<string> (res.rbegin(), res.rend());
    }
    void dfs(unordered_map<string, multiset<string>>& m, string s, vector<string>& res) {
        while (m[s].size()) {
            string t = *m[s].begin();
            m[s].erase(m[s].begin());
            dfs(m, t, res);
        }
        res.push_back(s);
    }
};

 

下面咱們來看迭代的解法，須要藉助棧來實現，來實現回溯功能。好比對下面這個例子：

tickets = [["JFK", "KUL"], ["JFK", "NRT"], ["MRT", "JFK"]]

那麼創建的圖以下：

JFK -> KUL, NRT

NRT -> JFK

因爲multiset是按順序存的，全部KUL會在NRT以前，那麼咱們起始從JFK開始遍歷，先到KUL，可是KUL沒有下家了，這時候圖中的邊並無遍歷完，此時咱們須要將KUL存入棧中，而後繼續往下遍歷，最後再把棧裏的節點存回結果便可，參見代碼以下：

 

解法二：

class Solution {
public:
    vector<string> findItinerary(vector<pair<string, string>> tickets) {
        vector<string> res;
        stack<string> st{{"JFK"}};
        unordered_map<string, multiset<string>> m;
        for (auto t : tickets) {
            m[t.first].insert(t.second);
        }
        while (!st.empty()) {
            string t = st.top(); 
            if (m[t].empty()) {
                res.insert(res.begin(), t);
                st.pop();
            } else {
                st.push(*m[t].begin());
                m[t].erase(m[t].begin());
            }
        }
        return res;
    }
};

 

相似題目：

Course Schedule

Course Schedule II

 

參考資料：

https://leetcode.com/problems/reconstruct-itinerary/

https://discuss.leetcode.com/topic/36370/short-ruby-python-java-c

https://discuss.leetcode.com/topic/36721/short-c-dfs-iterative-44ms-solution-with-explanation-no-recursive-calls-no-backtracking

 

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