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332. 重新安排行程

English Version

题目描述

给你一份航线列表 tickets ，其中 tickets[i] = [fromi, toi] 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"] 与 ["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
• fromi.length == 3
• toi.length == 3
• fromi 和 toi 由大写英文字母组成
• fromi != toi

解法

Python3

Java

class Solution {
    void dfs(Map<String, Queue<String>> adjLists, List<String> ans, String curr) {
        Queue<String> neighbors = adjLists.get(curr);
        if (neighbors == null) {
            ans.add(curr);
            return;
        }
        while (!neighbors.isEmpty()) {
            String neighbor = neighbors.poll();
            dfs(adjLists, ans, neighbor);
        }
        ans.add(curr);
        return;
    }

    public List<String> findItinerary(List<List<String>> tickets) {
        Map<String, Queue<String>> adjLists = new HashMap<>();
        for (List<String> ticket : tickets) {
            String from = ticket.get(0);
            String to = ticket.get(1);
            if (!adjLists.containsKey(from)) {
                adjLists.put(from, new PriorityQueue<>());
            }
            adjLists.get(from).add(to);
        }
        List<String> ans = new ArrayList<>();
        dfs(adjLists, ans, "JFK");
        Collections.reverse(ans);
        return ans;
    }
}

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