【VIP】【图】332. Reconstruct Itinerary【M】【47】-优快云博客

本文介绍了一个算法问题，即根据给定的机票对（包含出发地和目的地）重构旅行路线。该路线必须从JFK机场开始，并且需要找到字典序最小的合法行程。通过示例展示了如何解决这一问题。

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:
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.

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class Solution(object):

    def findItinerary(self, tickets):
        res = ['JFK']
        dic = collections.defaultdict(list)
        for flight in tickets:
            dic[flight[0]] += flight[1],

        #print dic

        self.dfs(dic,'JFK',res,len(tickets))
        return res

    def dfs(self,dic,leave,res,flights):
        if len(res) == flights + 1:
            return res
        currentDst = sorted(dic[leave]) #当前这个点对应的所有目的地

        for dst in currentDst:
            dic[leave].remove(dst)
            res.append(dst)

            valid = self.dfs(dic,dst,res,flights)
            if valid:
                return valid
            res.pop()
            dic[leave].append(dst)