332. 重新安排行程

给你一份航线列表tickets，其中tickets[i] = [from_i, to_i]表示飞机出发和降落的机场地点。请你对该行程进行重新规划排序。

所有这些机票都属于一个从JFK（肯尼迪国际机场）出发的先生，所以该行程必须从JFK开始。如果存在多种有效的行程，请你按字典排序返回最小的行程组合。

例如，行程["JFK", "LGA"]与["JFK", "LGB"]相比就更小，排序更靠前。
假定所有机票至少存在一种合理的行程。且所有的机票 必须都用一次 且 只能用一次。

示例 1：

输入：tickets = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
输出：["JFK","MUC","LHR","SFO","SJC"]

示例 2：

输入：tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
输出：["JFK","ATL","JFK","SFO","ATL","SFO"]
解释：另一种有效的行程是 ["JFK","SFO","ATL","JFK","ATL","SFO"] ，但是它字典排序更大更靠后。

提示：

• 1 <= tickets.length <= 300
• tickets[i].length == 2
• from_i.length == 3
• to_i.length == 3
• from_i和to_i由大写英文字母组成
• from_i != to_i

1.哈希+回溯
Python解答：

class Solution:
    def findItinerary(self, tickets: List[List[str]]) -> List[str]:
        adic, bdic = {}, {}
        for item in tickets:
            if item[0] not in adic.keys():
                adic[item[0]] = [item[1]]
            else:
                adic[item[0]].append(item[1])
            if (item[0]+item[1]) not in bdic.keys():
                bdic[item[0]+item[1]] = 1
            else:
                bdic[item[0]+item[1]] += 1
        lens = len(tickets)
        for key in adic.keys():
            adic[key].sort()
        result = None
        def backtrace(orig, res, i):
            nonlocal result
            if i == lens:
                result = copy.deepcopy(res)
                return 
            else:
                if orig in adic.keys():
                    dests = adic[orig]
                    for dest in dests:
                        if bdic[orig+dest] > 0:
                            bdic[orig+dest] -= 1
                            if not result:
                                backtrace(dest, res+[dest], i+1)
                            else:
                                return 
                            bdic[orig+dest] += 1

        backtrace('JFK', ['JFK'], 0)
        return result

2.Hierholzer算法
Python解答：

class Solution:
    def findItinerary(self, tickets: List[List[str]]) -> List[str]:
        adic = {}
        for item in tickets:
            if item[0] not in adic.keys():
                adic[item[0]] = [item[1]]
            else:
                adic[item[0]].append(item[1])
        for key in adic.keys():
            adic[key].sort()
        def dfs(node):
            if node in adic.keys():
                while adic[node]:
                    temp = adic[node].pop(0)
                    dfs(temp)
            stack.append(node)

        stack = []
        dfs('JFK')
        return stack[::-1]