[leetcode] 332. Reconstruct Itinerary @ python

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原题

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:

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”].
All airports are represented by three capital letters (IATA code).
You may assume all tickets form at least one valid itinerary.
Example 1:

Input: [[“MUC”, “LHR”], [“JFK”, “MUC”], [“SFO”, “SJC”], [“LHR”, “SFO”]]
Output: [“JFK”, “MUC”, “LHR”, “SFO”, “SJC”]
Example 2:

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.

解法

参考: 花花酱 LeetCode 332. Reconstruct Itinerary
字典 + DFS. 遍历tickets, 构造起点 - 终点的字典, 然后对字典的值进行倒序排列, 目的是使用graph[airport].pop()时, 获得的是排序最低的字符串, 然后进行深度优先搜索.

代码

class Solution(object):
    def findItinerary(self, tickets):
        """
        :type tickets: List[List[str]]
        :rtype: List[str]
        """
        def dfs(graph, airport, res):            
            while graph[airport]:
                next = graph[airport].pop()
                dfs(graph, next, res)
            res.append(airport)            
                
        graph = collections.defaultdict(list)
        for frm, to in tickets:
            graph[frm].append(to)
        for key in graph:
            graph[key].sort(reverse = True)
        res = []
        dfs(graph, 'JFK', res)
        return res[::-1]