From 5fbb5fdc2ca45d5e210f4f7a043a9da52ca1a1d3 Mon Sep 17 00:00:00 2001
From: David Gao <gao.ten@northeastern.edu>
Date: Sat, 13 Jan 2024 20:44:21 -0500
Subject: [PATCH] =?UTF-8?q?0322.=E9=87=8D=E6=96=B0=E5=AE=89=E6=8E=92?=
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---
 problems/0332.重新安排行程.md | 49 +++++++++++++++--------------
 1 file changed, 26 insertions(+), 23 deletions(-)

diff --git a/problems/0332.重新安排行程.md b/problems/0332.重新安排行程.md
index 1473ed05..036cfc51 100644
--- a/problems/0332.重新安排行程.md
+++ b/problems/0332.重新安排行程.md
@@ -449,34 +449,37 @@ class Solution:
 ```
 回溯 使用字典
 ```python	
-from collections import defaultdict
-
 class Solution:
     def findItinerary(self, tickets: List[List[str]]) -> List[str]:
-        targets = defaultdict(list)  # 构建机场字典
-        for ticket in tickets:
-            targets[ticket[0]].append(ticket[1])
-        for airport in targets:
-            targets[airport].sort()  # 对目的地列表进行排序
+        self.adj = {}
 
-        path = ["JFK"]  # 起始机场为"JFK"
-        self.backtracking(targets, path, len(tickets))
-        return path
+        # 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])
 
-    def backtracking(self, targets, path, ticketNum):
-        if len(path) == ticketNum + 1:
-            return True  # 找到有效行程
+        # 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]
 
-        airport = path[-1]  # 当前机场
-        destinations = targets[airport]  # 当前机场可以到达的目的地列表
-        for i, dest in enumerate(destinations):
-            targets[airport].pop(i)  # 标记已使用的机票
-            path.append(dest)  # 添加目的地到路径
-            if self.backtracking(targets, path, ticketNum):
-                return True  # 找到有效行程
-            targets[airport].insert(i, dest)  # 回溯，恢复机票
-            path.pop()  # 移除目的地
-        return False  # 没有找到有效行程
+        # 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
 
 ```
 回溯 使用字典 逆序