Question:
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:
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"]
.
All airports are represented by three capital letters (IATA code).
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”]
Solution 1: Iteration by using stack
class Solution {
public List<String> findItinerary(List<List<String>> tickets) {
Map<String, PriorityQueue<String>> map = new HashMap<>();
List<String> route = new ArrayList<>();
for (List<String> ticket : tickets) {
map.computeIfAbsense(ticket.get(0), k -> new PriorityQueue()).add(ticket.get(1));
}
Stack<Spring> stack = new Stack<>();
stack.push("JFK");
while(!stack.isEmpty()) { // here we must put the while loop as the
// out loop
while(map.containsKey(stack.peek()) && !map.get(stack.peek()).isEmpty()) {
stack.push(map.get(stack.peek()).poll());
}
route.add(0, stack.pop());
}
return route;
}
}
Solution 2: Recursive Solution
class Solution {
List<String> route = new ArrayList<>();
Map<String, List<String>> map = new HashMap<>();
public List<String> findItinerary(List<List<String>> tickets) {
for (List<String> ticket : tickets) {
map.countIfAbsense(ticket.get(0), k -> new PriorityQueue()).add(ticket.get(1));
}
visit("JFK");
return route;
}
public void visit(String airport) {
while (map.containsKey(airport) && !map.get(airport).isEmpty()) {
visit(map.get(airport).poll());
}
route.add(0, airport);
}
}
Note:
Pay attention to the conditions in the while loop.