[LeetCode] Reconstruct Itinerary 重建行程单

 

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.

 

这道题给咱们一堆飞机票，让咱们创建一个行程单，若是有多种方法，取其中字母顺序小的那种方法。这道题的本质是有向图的遍历问题，那么LeetCode关于有向图的题只有两道Course Schedule和Course Schedule II，而那两道是关于有向图的顶点的遍历的，而本题是关于有向图的边的遍历。每张机票都是有向图的一条边，咱们须要找出一条通过全部边的路径，那么DFS不是咱们的不二选择。先来看递归的结果，咱们首先把图创建起来，经过邻接链表来创建。因为题目要求解法按字母顺序小的，那么咱们考虑用multiset，能够自动排序。等咱们图创建好了之后，从节点JFK开始遍历，只要当前节点映射的multiset里有节点，咱们取出这个节点，将其在multiset里删掉，而后继续递归遍历这个节点，因为题目中限定了必定会有解，那么等图中全部的multiset中都没有节点的时候，咱们把当前节点存入结果中，而后再一层层回溯回去，将当前节点都存入结果，那么最后咱们结果中存的顺序和咱们须要的相反的，咱们最后再翻转一下便可，参见代码以下：

 

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);
    }
};

 

下面咱们来看迭代的解法，须要借助栈来实现，来实现回溯功能。好比对下面这个例子：

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

那么创建的图以下：

JFK -> KUL, NRT

NRT -> JFK

因为multiset是按顺序存的，全部KUL会在NRT以前，那么咱们起始从JFK开始遍历，先到KUL，可是KUL没有下家了，这时候图中的边并无遍历完，此时咱们须要将KUL存入栈中，而后继续往下遍历，最后再把栈里的节点存回结果便可，参见代码以下：

 

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;
    }
};

 

相似题目：

Course Schedule

Course Schedule II

 

参考资料：

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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