332. 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:
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:

tickets = [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Return ["JFK", "MUC", "LHR", "SFO", "SJC"].

Example 2:

tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
Return ["JFK","ATL","JFK","SFO","ATL","SFO"].
Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"]. But it is larger in lexical order.

Solution：DFS "一笔找Route"

思路： 类似拓扑排序Topological Sort DFS思路
区别在于是否可以有loop(这题可以有，所以不用check visisted)
Topological Sort DFS: http://www.jianshu.com/p/5880cf3be264
可参考210题 Course Schedule II solution2: http://www.jianshu.com/p/96841bf6f167

因为node是string，所以adj_list可以用map实现。
实现1_a: 结果存在linkedlist；
实现1_b: 结果存在stack；
其实目的是相同的。

Time Complexity: O(E) Space Complexity: O(E)

Solution1 Code：

class Solution {
    Map> adj_map = new HashMap<>();
    List route = new LinkedList<>();
    
    
    public List findItinerary(String[][] tickets) {
        route = new LinkedList<>();
        adj_map = new HashMap<>();
        
        for (String[] ticket : tickets)
            adj_map.computeIfAbsent(ticket[0], k -> new PriorityQueue()).add(ticket[1]);
        dfs("JFK");
        return route;
    }

    private void dfs(String airport) {
        while(adj_map.containsKey(airport) && !adj_map.get(airport).isEmpty()) {
            String des = adj_map.get(airport).poll();
            dfs(des);
        }
        route.add(0, airport);
    }
}

Solution1b Code：

class Solution {
    Map> adj_map = new HashMap<>();
    Deque stack;
    
    
    public List findItinerary(String[][] tickets) {
        stack = new ArrayDeque<>();
        adj_map = new HashMap<>();
        
        for (String[] ticket : tickets)
            adj_map.computeIfAbsent(ticket[0], k -> new PriorityQueue()).add(ticket[1]);
        dfs("JFK");
        return new ArrayList(stack);
    }

    private void dfs(String airport) {
        while(adj_map.containsKey(airport) && !adj_map.get(airport).isEmpty()) {
            String des = adj_map.get(airport).poll();
            dfs(des);
        }
        stack.push(airport);
    }
}

Round1

class Solution {
    public List findItinerary(String[][] tickets) {
        if(tickets == null || tickets.length == 0) return new ArrayList<>();
        
        Map> adj_list = new HashMap<>();
        Deque stack = new ArrayDeque<>();
        
        // graph init
        for(String[] ticket: tickets) {
            if(!adj_list.containsKey(ticket[0])) {
                adj_list.put(ticket[0], new PriorityQueue<>());
            }
            adj_list.get(ticket[0]).offer(ticket[1]);
        }
        
        dfs(adj_list, "JFK", stack);
        
        return new ArrayList<>(stack);   
    }
    
    private void dfs(Map> adj_list, String cur, Deque stack) {
        if(adj_list.get(cur) != null) {
            while(!adj_list.get(cur).isEmpty()) {
                String next = adj_list.get(cur).poll();
                dfs(adj_list, next, stack);
            }
        }
        stack.push(cur);
    }
}