README_CN.md

332. 重新安排行程

给你一份航线列表 tickets ，其中 tickets[i] = [fromi, toi] 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
• fromi.length == 3
• toi.length == 3
• fromi 和 toi 由大写英文字母组成
• fromi != toi

题解 (Python)

1. 题解

class Solution:
    def findItinerary(self, tickets: List[List[str]]) -> List[str]:
        targets = {}
        stack = []
        used = set()
        itinerary = [("JFK", None, None)]

        for fr0m, to in tickets:
            if fr0m not in targets:
                targets[fr0m] = []
            if to not in targets:
                targets[to] = []
            targets[fr0m].append(to)
        for fr0m in targets:
            targets[fr0m].sort(reverse=True)

        while True:
            fr0m = itinerary[-1][0]

            for i in range(len(targets[fr0m])):
                if (fr0m, i) not in used:
                    if stack == [] \
                            or fr0m != stack[-1][0] \
                            or targets[fr0m][i] != targets[fr0m][stack[-1][1]] \
                            or len(itinerary) != stack[-1][2]:
                        stack.append((fr0m, i, len(itinerary)))

            while stack[-1][2] < len(itinerary):
                _, fr0m, i = itinerary.pop()
                used.remove((fr0m, i))

            fr0m, i, _ = stack.pop()
            used.add((fr0m, i))
            itinerary.append((targets[fr0m][i], fr0m, i))

            if len(itinerary) == len(tickets) + 1:
                return [airport for airport, _, _ in itinerary]