【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
考虑考虑1 小时前
JDK9中的dropWhile
java·后端·java ee
想躺平的咸鱼干1 小时前
Volatile解决指令重排和单例模式
java·开发语言·单例模式·线程·并发编程
hqxstudying1 小时前
java依赖注入方法
java·spring·log4j·ioc·依赖
·云扬·1 小时前
【Java源码阅读系列37】深度解读Java BufferedReader 源码
java·开发语言
Tanecious.2 小时前
LeetCode 876. 链表的中间结点
算法·leetcode·链表
Bug退退退1233 小时前
RabbitMQ 高级特性之重试机制
java·分布式·spring·rabbitmq
小皮侠3 小时前
nginx的使用
java·运维·服务器·前端·git·nginx·github
Zz_waiting.3 小时前
Javaweb - 10.4 ServletConfig 和 ServletContext
java·开发语言·前端·servlet·servletconfig·servletcontext·域对象
全栈凯哥3 小时前
02.SpringBoot常用Utils工具类详解
java·spring boot·后端
兮动人3 小时前
获取终端外网IP地址
java·网络·网络协议·tcp/ip·获取终端外网ip地址