332. Reconstruct Itinerary

https://leetcode.com/problems/reconstruct-itinerary/

Given a list of airline tickets represented by pairs of departure and arrival airports [from, to], reconstruct the itinerary in order. All of the tickets belong to a man who departs from JFK. Thus, the itinerary must begin with JFK.

Note:

1. 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"].
2. All airports are represented by three capital letters (IATA code).
3. You may assume all tickets form at least one valid itinerary.

Example 1:

Input: [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]

Example 2:

Input: [["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.

--- ---

	class Solution {
	public List<String> findItinerary(List<List<String>> tickets) {
	List<String> ans = new ArrayList<String>();
	// Priority Queue to order lexicographically, and remove once used
	Map<String, PriorityQueue<String>> graph = new HashMap<String, PriorityQueue<String>>();
	buildGraph(tickets, graph);
	// Post Order DFS => topological sort
	dfs(graph, "JFK", ans);
	return ans;
	}
	private void dfs(Map<String, PriorityQueue<String>> graph, String u, List<String> ans) {
	// Greedy DFS .. no backtracking
	// Note - Repeat expression graph.get(u),
	// Do not use variable for graph.get(u), because PQ.poll() empties the graph
	while (graph.get(u) != null && !graph.get(u).isEmpty()) {
	dfs(graph, graph.get(u).poll(), ans);
	}
	// Post Order DFS
	// Add first element to avoid Collections reverse in the end
	ans.add(0, u);
	}
	private void buildGraph(List<List<String>> tickets, Map<String, PriorityQueue<String>> graph) {
	for (List<String> edge: tickets) {
	String u = edge.get(0);
	String v = edge.get(1);
	graph.putIfAbsent(u, new PriorityQueue<String>());
	graph.get(u).offer(v);
	}
	}
	}

view raw 332.ReconstructItineraryDFSPostOrder.java hosted with ❤ by GitHub

	class Solution {
	public List<String> findItinerary(List<List<String>> tickets) {
	List<String> ans = new ArrayList<String>();
	if (tickets == null || tickets.size() == 0) {
	return ans;
	}
	Map<String, PriorityQueue<String>> graph = new HashMap<String, PriorityQueue<String>>();
	for (List<String> trip: tickets) {
	String from = trip.get(0);
	String to = trip.get(1);
	PriorityQueue<String> destination = graph.getOrDefault(from, new PriorityQueue<String>());
	destination.add(to);
	graph.put(from, destination);
	}
	Stack<String> stack = new Stack<String>();
	// Starting node
	stack.push("JFK");
	// Nodes still to be visited
	while (!stack.isEmpty()) {
	// Destination nodes exist
	while (graph.get(stack.peek()) != null && !graph.get(stack.peek()).isEmpty()) {
	// Priority Queue - try min lexicographical first, and discard after use
	stack.push(graph.get(stack.peek()).poll());
	}
	// Add to first position to avoid Collections reverse in the end
	ans.add(0, stack.pop());
	}
	return ans;
	}
	}

view raw 332.ReconstructItineraryIterative.java hosted with ❤ by GitHub