항해99 2주차 - 일정재구성

Today I learned
2022/01/20

회고록

1/20

항해 99, 알고리즘 1주차

교재 : 파이썬 알고리즘 인터뷰

12장 그래프

1. 이론

https://velog.io/@jsw4215/%ED%95%AD%ED%95%B499-2%EC%A3%BC%EC%B0%A8-%EA%B7%B8%EB%9E%98%ED%94%84DFSBFS-%EC%9D%B4%EB%A1%A0-%EC%A0%95%EB%A6%AC

2. 문제

You are given a list of airline tickets where tickets[i] = [fromi, toi] represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.

All of the tickets belong to a man who departs from "JFK", thus, the itinerary must begin with "JFK". If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string.

For example, the itinerary ["JFK", "LGA"] has a smaller lexical order than ["JFK", "LGB"].
You may assume all tickets form at least one valid itinerary. You must use all the tickets once and only once.

Example 1:

Input: tickets = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
Output: ["JFK","MUC","LHR","SFO","SJC"]

Example 2:

Input: tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
Output: ["JFK","ATL","JFK","SFO","ATL","SFO"]
Explanation: Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"] but it is larger in lexical order.

Constraints:

1 <= tickets.length <= 300
tickets[i].length == 2
fromi.length == 3
toi.length == 3
fromi and toi consist of uppercase English letters.
fromi != toi

https://leetcode.com/problems/reconstruct-itinerary/

3. MySol

• Recursive DFS

import collections


def solution(schedules):

    arranged = collections.defaultdict(list)

    for a,b in sorted(schedules):
        arranged[a].append(b)

    curr_node = []
    next_node = []
    result = []

    def connection(dep):
        while arranged[dep]:
            connection(arranged[dep].pop(0))
        result.append(dep)

    connection('JFK')


    return result[::-1]


if __name__ == '__main__':

    schedules = [["JFK","KUL"],["JFK","NRT"],["NRT","JFK"]]
    #[["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
    #[["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]

    result = solution(schedules)

    print('result : ' + str(result))

4. 배운 점

• 해당 문제는 이해하게 되면 DFS및 재귀함수에 대해 이해하는데 굉장히 큰 도움이 되는 문제이다.

5. 총평

재귀, DFS 훈련