【陪伴式刷题】Day 26|回溯|332.重新安排行程(Reconstruct Itinerary)

刷题顺序按照代码随想录建议

题目描述

英文版描述

You are given a list of airline tickets where tickets[i] = [from(i), to(i)] 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
  • from(i).length == 3
  • to(i).length == 3
  • from(i) and to(i) consist of uppercase English letters.
  • from(i) != to(i)

英文版地址

leetcode.com/problems/re...

中文版描述

给你一份航线列表 tickets ,其中 tickets[i] = [from(i), to(i)] 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"]["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
  • from(i).length == 3
  • to(i).length == 3
  • from(i)to(i) 由大写英文字母组成
  • from(i) != to(i)

中文版地址

leetcode.cn/problems/re...

解题方法

递归法

Java 复制代码
class Solution {
    List<String> result = new ArrayList<>();
    Map<String, TreeMap<String, Integer>> map = new HashMap<>();

    public List<String> findItinerary(List<List<String>> tickets) {
        // 初始化map
        for (List<String> ticket : tickets) {
            String from = ticket.get(0);
            String end = ticket.get(1);
            map.putIfAbsent(from, new TreeMap<>());
            TreeMap<String, Integer> fromTreeMap = map.get(from);
            fromTreeMap.put(end, fromTreeMap.getOrDefault(end, 0) + 1);
        }
        result.add("JFK");
        boolean flag = backTrack(tickets, 0);
        return result;
    }

    private boolean backTrack(List<List<String>> tickets, int count) {
        if (result.size() == tickets.size() + 1) {
            return true;
        }
        String from = result.get(result.size() - 1);
        TreeMap<String, Integer> stringIntegerTreeMap = map.get(from);
        if(stringIntegerTreeMap == null){
            return false;
        }
        for (String ss : stringIntegerTreeMap.keySet()) {
            Integer mm = stringIntegerTreeMap.get(ss);
            if (mm > 0) {
                result.add(ss);
                map.get(from).put(ss, stringIntegerTreeMap.get(ss) - 1);
                boolean b = backTrack(tickets, count + 1);
                if (b) {
                    return true;
                }
                result.remove(result.size() - 1);
                map.get(from).put(ss, stringIntegerTreeMap.get(ss) + 1);
            }
        }
        return false;
    }
}

复杂度分析

  • 时间复杂度:不会啊不会啊= =// >> 回溯算法的时间复杂度,搞不懂啊(q求大神指点=,,_,,:3)
  • 空间复杂度:O(n),为递归过程中栈的开销,平均情况下为 O(log⁡n),最坏情况下树呈现链状,为 O(n)
相关推荐
karry_k4 小时前
MyBatis批量insert-select踩坑:useGeneratedKeys=true 可能让PostgreSQL返回大量插入结果
java·后端
karry_k4 小时前
PostgreSQL 在 MyBatis 中执行正常 SQL 失效:一次 DELETE USING 踩坑记录
java·后端
SamDeepThinking8 小时前
从源码到代码:MyBatis-Flex 与 MyBatis-Plus 的逐项对比
java·后端·程序员
她的男孩11 小时前
Spring Boot 接 Flowable 工作流:用 3 个注解搭一个请假审批流程
java·后端·架构
荣码12 小时前
LLM结构化输出:让AI返回JSON而不是废话,我踩了4个坑
java·python
plainGeekDev14 小时前
Gson → kotlinx.serialization
android·java·kotlin
小bo波1 天前
Java Swing 图形用户界面实验 —— 从算术练习到游戏开发的完整实践
java·课程设计·gui·游戏开发·扫雷·swing
咖啡八杯1 天前
GoF设计模式——备忘录模式
java·后端·spring·设计模式