• 【LeetCode】332. Reconstruct Itinerary


    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:

    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 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.

    提示:

    这道题看到之后第一反应一般就是使用dfs,不过用法还需要结合题目的要求和图的特性。因为题目要求结果要按照字母大小顺序排序,因此我们在记录每个点能到达的点集时,可以用STL中的unordered_map<string, multiset<string>>, unordered_map效率比map高一些,另外multiset允许同样的值插入多次,且是按序插入的。

    然后说说这个图的特性,因为题目说了给定的输入是一定有解的,所以,图中所有的点我们可以按照出度和入度之和的奇偶分成两类:

    • 出度和入度之和为奇:这种点最多只有两个,就是起点和终点;
    • 出度和入度之和为偶:就是正常的中间过渡点;
    • 如果所有点的出度和入度之和都为偶,那么一直dfs到底就是要求的解;
    • 在dfs过程中,如果我们stuck了,其实就是因为我们访问到了终点。

    上面这几个特性就不一一证明了,可以画个草图简单理解一下。

    因此我们在dfs的时候,如果卡住了,那么说明访问到了终点,就把这个点放进vector中。如果没卡住的话,就把点push进stack中(用于回溯),并且一直访问下去,并且经过的点都要记得及时删除,防止走重复的路径。

    最后,由于先访问到的“终点”在vector的前端,因此在返回vector前要记得reverse一下。

    代码:

    class Solution {
    public:
        vector<string> findItinerary(vector<pair<string, string>> tickets) {
            unordered_map<string, multiset<string>> m;
            vector<string> res;
            if (tickets.size() <= 0) {
                return res;
            }
            for (pair<string, string> p: tickets) {
                m[p.first].insert(p.second);
    
            }
            stack<string> s;
            s.push("JFK");
            while (s.size()) {
                string next = s.top();
                if (m[next].empty()) {
                    res.push_back(next);
                    s.pop();
                } else {
                    s.push(*m[next].begin());
                    m[next].erase(m[next].begin());
                }
            }
            reverse(res.begin(), res.end());
            return res;
        }
    };
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  • 原文地址:https://www.cnblogs.com/jdneo/p/5318509.html
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