【leetcode】332. Reconstruct Itinerary

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:

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:

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