Reconstruct Itinerary(Medium)

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.

解题思路1：欧拉通路（Eulerian path）

将机场视为顶点，机票视为看做有向边，可以构成一个有向图。

通过图（无向图或有向图）中所有边且每边仅通过一次的通路称为欧拉通路，相应的回路称为欧拉回路。具有欧拉回路的图称为欧拉图（Euler Graph），具有欧拉通路而无欧拉回路的图称为半欧拉图。

因此题目的实质就是从JFK顶点出发寻找欧拉通路，可以利用Hierholzer算法。

public List<String> findItinerary(String[][] tickets) {
    Map<String, PriorityQueue<String>> targets = new HashMap<>();
    for (String[] ticket : tickets)
        targets.computeIfAbsent(ticket[0], k -> new PriorityQueue()).add(ticket[1]);

    List<String> route = new LinkedList();
    Stack<String> stack = new Stack<>();
    stack.push("JFK");
    while (!stack.empty()) {
        String airport = stack.peek();
        while (targets.containsKey(airport) && !targets.get(airport).isEmpty()){
            stack.push( targets.get(airport).poll() );
            airport = stack.peek();
        }

        route.add(0, stack.pop());
    }
    return route;
}


public static void main(String[] args) {
    Test instance = new Test();

    //Return ["JFK","ATL","JFK","SFO","ATL","SFO"]
    String[][] tickets = {{"JFK","SFO"},{"JFK","ATL"},{"SFO","ATL"},{"ATL","JFK"},{"ATL","SFO"}};
    int[] nums = {4,1,1,2,2};

    System.out.println("doing... " + instance.findItinerary(tickets));
}

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解题思路2：DFS+Backtracking

public List<String> findItinerary(String[][] tickets) {
    ArrayList<String> result = new ArrayList<String>();

    if(tickets == null || tickets.length == 0){
        return result;
    }

    int total = tickets.length + 1;

    HashMap<String, ArrayList<String>> map = new HashMap<String, ArrayList<String>>();

    for(int i = 0; i < tickets.length; i++){
        if(map.containsKey(tickets[i][0])){
            ArrayList<String> tmp = map.get(tickets[i][0]);
            listAdd(tickets[i][1], tmp);
        } else{
            ArrayList<String> tmp = new ArrayList<String>();
            tmp.add(tickets[i][1]);
            map.put(tickets[i][0], tmp);
        }
    }

    result.add("JFK");

    itineraryHelper("JFK", map, result, total, 1);

    return result;
}

public boolean itineraryHelper(String current, HashMap<String, ArrayList<String>> map,
                                   ArrayList<String> result, int total, int num){

    if(num >= total){
        return true;
    }

    if(!map.containsKey(current) || map.get(current).size() == 0){
        return false;
    }

    ArrayList<String> curList = map.get(current);
    int i = 0;

    while(i < curList.size()){
        String next = curList.remove(i);
        result.add(next);

        if(itineraryHelper(next, map, result, total, num + 1)){
            return true;
        }

        result.remove(result.size() - 1);
        listAdd(next, curList);
        i++;
    }

    return false;
}

public void listAdd(String value, ArrayList<String> list){
    if(list.size() == 0){
        list.add(value);
        return;
    } else{
        int i = 0;
        while(i < list.size()){
            if(value.compareTo(list.get(i)) <= 0){
                list.add(i, value);
                return;
            }
            i++;
        }
        list.add(value);
        return;
    }
}

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