feat: solve No.332

Author: BaffinLee

301-400/332. Reconstruct Itinerary.md

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+# 332. Reconstruct Itinerary
+
+- Difficulty: Hard.
+- Related Topics: Depth-First Search, Graph, Eulerian Circuit.
+- Similar Questions: Longest Common Subpath, Valid Arrangement of Pairs.
+
+## Problem
+
+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:
+
+![](https://assets.leetcode.com/uploads/2021/03/14/itinerary1-graph.jpg)
+
+```
+Input: tickets = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
+Output: ["JFK","MUC","LHR","SFO","SJC"]
+```
+
+Example 2:
+
+![](https://assets.leetcode.com/uploads/2021/03/14/itinerary2-graph.jpg)
+
+```
+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`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {string[][]} tickets
+ * @return {string[]}
+ */
+var findItinerary = function(tickets) {
+    tickets.sort((a, b) => (a[1] === b[1] ? 0 : (a[1] < b[1] ? -1 : 1)));
+
+    var map = {};
+    for (var i = 0; i < tickets.length; i++) {
+      if (!map[tickets[i][0]]) map[tickets[i][0]] = [];
+      map[tickets[i][0]].push(tickets[i][1]);
+    }
+
+    console.log(map, tickets);
+
+  var itinerary = [];
+  dfs('JFK', map, itinerary);
+  return itinerary.reverse();
+};
+
+var dfs = function(airport, map, itinerary) {
+  while (map[airport]?.length) {
+    dfs(map[airport].shift(), map, itinerary);
+  }
+  itinerary.push(airport);
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n * log(n)).
+* Space complexity : O(n).