【leetcode】332. Reconstruct Itinerary

机票行程重构算法解析

最新推荐文章于 2022-04-27 16:18:21 发布

转载最新推荐文章于 2022-04-27 16:18:21 发布 · 100 阅读

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原文链接：http://www.cnblogs.com/seyjs/p/10107050.html

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#数据结构与算法

本文详细解析了一道经典的算法题目，即如何从一系列由出发地和目的地组成的航班票务信息中，重构出从JFK出发的完整行程路径，并确保路径结果的字典序最小。文章提供了详细的解题思路，包括如何利用欧拉回路的概念解决此问题，同时考虑到多条相同起点和终点的路径情况。并通过具体示例展示了算法的实现过程。

题目如下：

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.

解题思路：本题是非常典型的欧拉回路的场景，但有两点需要注意的，一是要求返回结果是最小字典序，所有事先需要对所有机场先按升序排好；第二点有点坑，欧拉回路一条边只能走一次，但本题中输入的tickets中，起点和终点相同的ticket可能有多个。

代码如下：

class Solution(object):
    res = []
    def euler(self,airports,board,portlist):
        for i in range(len(board[airports])):
            if board[airports][i] > 0 :
                board[airports][i] -= 1
                self.euler(i,board,portlist)
        self.res.insert(0,portlist[airports])


    def findItinerary(self, tickets):
        """
        :type tickets: List[List[str]]
        :rtype: List[str]
        """
        #print len(tickets)
        self.res = []
        inx = 0
        dic = {}
        for (v1,v2) in tickets:
            if v1 not in dic:
                dic[v1] = 1
            if v2 not in dic:
                dic[v2] = inx

        portlist = dic.keys()
        portlist.sort()

        board = []
        dic = {}
        for i in range(len(portlist)):
            board.append([0]*len(portlist))
            dic[portlist[i]] = i

        for (v1, v2) in tickets:
            board[dic[v1]][dic[v2]] += 1
            
        self.euler(dic['JFK'],board,portlist)
        return self.res