reconstruct-itinerary

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Algorithms

You are given a list of airline <code>tickets</code> where <code>tickets[i] = [fromi, toi]</code> 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 <code>&quot;JFK&quot;</code>, thus, the itinerary must begin with <code>&quot;JFK&quot;</code>. If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string.

<ul>
	<li>For example, the itinerary <code>[&quot;JFK&quot;, &quot;LGA&quot;]</code> has a smaller lexical order than <code>[&quot;JFK&quot;, &quot;LGB&quot;]</code>.</li>
</ul>

You may assume all tickets form at least one valid itinerary. You must use all the tickets once and only once.

&nbsp;
Example 1:
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/14/itinerary1-graph.jpg" style="width: 382px; height: 222px;" />
<pre>
Input: tickets = [[&quot;MUC&quot;,&quot;LHR&quot;],[&quot;JFK&quot;,&quot;MUC&quot;],[&quot;SFO&quot;,&quot;SJC&quot;],[&quot;LHR&quot;,&quot;SFO&quot;]]
Output: [&quot;JFK&quot;,&quot;MUC&quot;,&quot;LHR&quot;,&quot;SFO&quot;,&quot;SJC&quot;]
</pre>

Example 2:
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/14/itinerary2-graph.jpg" style="width: 222px; height: 230px;" />
<pre>
Input: tickets = [[&quot;JFK&quot;,&quot;SFO&quot;],[&quot;JFK&quot;,&quot;ATL&quot;],[&quot;SFO&quot;,&quot;ATL&quot;],[&quot;ATL&quot;,&quot;JFK&quot;],[&quot;ATL&quot;,&quot;SFO&quot;]]
Output: [&quot;JFK&quot;,&quot;ATL&quot;,&quot;JFK&quot;,&quot;SFO&quot;,&quot;ATL&quot;,&quot;SFO&quot;]
Explanation: Another possible reconstruction is [&quot;JFK&quot;,&quot;SFO&quot;,&quot;ATL&quot;,&quot;JFK&quot;,&quot;ATL&quot;,&quot;SFO&quot;] but it is larger in lexical order.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= tickets.length &lt;= 300</code></li>
	<li><code>tickets[i].length == 2</code></li>
	<li><code>fromi.length == 3</code></li>
	<li><code>toi.length == 3</code></li>
	<li><code>fromi</code> and <code>toi</code> consist of uppercase English letters.</li>
	<li><code>fromi != toi</code></li>
</ul>

findItinerary

Reconstruct Itinerary

Longest Common Subpath

longest-common-subpath

最长公共子路径

Valid Arrangement of Pairs

valid-arrangement-of-pairs

合法重新排列数对

Depth-First Search

Graph

Eulerian Circuit

给你一份航线列表 <code>tickets</code> ，其中 <code>tickets[i] = [fromi, toi]</code> 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 <code>JFK</code>（肯尼迪国际机场）出发的先生，所以该行程必须从 <code>JFK</code> 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。

<ul>
	<li>例如，行程 <code>["JFK", "LGA"]</code> 与 <code>["JFK", "LGB"]</code> 相比就更小，排序更靠前。</li>
</ul>

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：
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/14/itinerary1-graph.jpg" style="width: 382px; height: 222px;" />
<pre>
输入：tickets = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
输出：["JFK","MUC","LHR","SFO","SJC"]
</pre>

示例 2：
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/14/itinerary2-graph.jpg" style="width: 222px; height: 230px;" />
<pre>
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。
</pre>

 

提示：

<ul>
	<li><code>1 <= tickets.length <= 300</code></li>
	<li><code>tickets[i].length == 2</code></li>
	<li><code>fromi.length == 3</code></li>
	<li><code>toi.length == 3</code></li>
	<li><code>fromi</code> 和 <code>toi</code> 由大写英文字母组成</li>
	<li><code>fromi != toi</code></li>
</ul>