332.重新安排

class Solution:
    def findItinerary(self, tickets):
        targets = defaultdict(list)  # 创建默认字典，用于存储机场映射关系
        for ticket in tickets:
            targets[ticket[0]].append(ticket[1])  # 将机票输入到字典中
        
        for key in targets:
            targets[key].sort(reverse=True)  # 对到达机场列表进行字母逆序排序
        
        result = []
        self.backtracking("JFK", targets, result)  # 调用回溯函数开始搜索路径
        return result[::-1]  # 返回逆序的行程路径
    
    def backtracking(self, airport, targets, result):
        while targets[airport]:  # 当机场还有可到达的机场时
            next_airport = targets[airport].pop()  # 弹出下一个机场
            self.backtracking(next_airport, targets, result)  # 递归调用回溯函数进行深度优先搜索
        result.append(airport)  # 将当前机场添加到行程路径中


class Solution:
    def findItinerary(self, tickets: List[List[str]]) -> List[str]:
        self.adj = {}

        # sort by the destination alphabetically
        # 根据航班每一站的重点字母顺序排序
        tickets.sort(key=lambda x:x[1])

        # get all possible connection for each destination
        # 罗列每一站的下一个可选项
        for u,v in tickets:
            if u in self.adj: self.adj[u].append(v)
            else: self.adj[u] = [v]

        # 从JFK出发
        self.result = []
        self.dfs("JFK")  # start with JFK

        return self.result[::-1]  # reverse to get the result

    def dfs(self, s):
        # if depart city has flight and the flight can go to another city
        while s in self.adj and len(self.adj[s]) > 0:
            # 找到s能到哪里，选能到的第一个机场
            v = self.adj[s][0]  # we go to the 1 choice of the city
            # 在之后的可选项机场中去掉这个机场
            self.adj[s].pop(0)  # get rid of this choice since we used it
            # 从当前的新出发点开始
            self.dfs(v)  # we start from the new airport

        self.result.append(s)  # after append, it will back track to last node, thus the result list is in reversed order

332. 重新安排行程 - 力扣（LeetCode）