0332-重新安排行程
阿新 • • 發佈:2021-11-23
給你一份航線列表 tickets ,其中 tickets[i] = [fromi, toi] 表示飛機出發和降落的機場地點。請你對該行程進行重新規劃排序。
所有這些機票都屬於一個從 JFK(肯尼迪國際機場)出發的先生,所以該行程必須從 JFK 開始。如果存在多種有效的行程,請你按字典排序返回最小的行程組合。
例如,行程 ["JFK", "LGA"] 與 ["JFK", "LGB"] 相比就更小,排序更靠前。
假定所有機票至少存在一種合理的行程。且所有的機票 必須都用一次 且 只能用一次。
示例 1:
輸入:tickets = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
輸出:["JFK","MUC","LHR","SFO","SJC"]
示例 2:
輸入:tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
輸出:["JFK","ATL","JFK","SFO","ATL","SFO"]
解釋:另一種有效的行程是 ["JFK","SFO","ATL","JFK","ATL","SFO"] ,但是它字典排序更大更靠後。
提示:
1 <= tickets.length <= 300
tickets[i].length == 2
fromi.length == 3
toi.length == 3
fromi 和 toi 由大寫英文字母組成
fromi != toi
來源:力扣(LeetCode)
連結:https://leetcode-cn.com/problems/reconstruct-itinerary
參考:
python
# 0332.重新安排行程 class Solution: def findItinerary(self, tickets:[[str]]) -> [str]: from collections import defaultdict tickets_dict = defaultdict(list) for item in tickets: tickets_dict[item[0]].append(item[1]) path = ["JFK"] def track(start_point): # 終止條件 if len(path) == len(tickets) + 1: return True tickets_dict[start_point].sort() for _ in tickets_dict[start_point]: # 必須及時刪除,避免出現死迴圈 end_point = tickets_dict[start_point].pop(0) path.append(end_point) # 只要找到一個就可以返回 if track(end_point): return True path.pop() tickets_dict[start_point].append(end_point) track("JFK") return path
golang
待完善