LeetCode-332 Reconstruct Itinerary

题目描述

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.

题目大意

字符串数组中包含n个成对出现的字符串，表示从起始地到目的地的机票，要求将所有的机票行程安排起来，前后有序，以“JFK”为起始地，形成一个行程。

（默认一定会找到一个可行的行程，同时若有多个行程安排，按照字典序排列最小的优先）

示例

E1

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

E2

Input: [["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.

解题思路

基于DFS的思想，利用map来保存每班飞机的起始地到目的地，递归的过程中依次删除访问过的地点。

复杂度分析

时间复杂度：O(N)

空间复杂度：O(N)

代码

class Solution {
public:
    vector<string> findItinerary(vector<vector<string>>& tickets) {
        // 将tickets中的起始地与目的地保存在map中
        for(int i = 0; i < tickets.size(); ++i)
            airl[tickets[i][0]].insert(tickets[i][1]);
        // DFS递归
        dfs("JFK");
        // 由于是按逆序保存的，所以返回结果的逆序
        return vector<string>(res.rbegin(), res.rend());
    }
    
    void dfs(string occa) {
        // 若当前起始地的目的地不为0，则依次访问目的地，并更新目的地，同时递归
        while(airl[occa].size()) {
            string tmp = *airl[occa].begin();
            airl[occa].erase(airl[occa].begin());
            dfs(tmp);
        }
        
        res.push_back(occa);
    }
    
private:
    unordered_map<string, multiset<string> > airl;
    vector<string> res;
};