332. Reconstruct Itinerary （leetcode）

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:
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.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

开始以为是寻找最短路径，发现有很大区别，比如有重复节点等等。不过写到最后自己在idea上运行通过，还是花了一些时间：

其中用了异常来跳出递归，这也是在网上看到的一种跳出递归的方法，此前没有用过，学习了。

（我自己写遍历经常直接写拼音。。。这个习惯不太好，以后也会注意。。）

public class Solution {
    static List<String> xulie(String a,String[][] ti,int n){
        ArrayList L= new ArrayList();
        for(int i = 0;i<n;i++){
            if(ti[i][0].equals(a)){
                L.add(ti[i][1]);
            }
        }
        if(L.size()>1){
            Collections.sort(L);
        }
        
        return L;
    }
    static void bianli(String a,String[][] tickets,List<String> h,Map<String,Integer> map,int n,int dp,Map<String,String> mape){
        List<String> L= new ArrayList();
        int p =0;
       
            L = xulie(a,tickets,n);
            int k =L.size();
            for(int t=0;t<k;t++){
                String b = L.get(t);
                if(map.get(b)==null){
                    if(dp == n-1){
                         h.add(b);
                     throw new StopMsgException();
                    }else{
                        continue;
                    }
                }else{
                    p = (int)map.get(b);
                if(p==0 || mape.get(a).toString().equals(b)==false){
                    a = b;
                    h.add(a);
                     map.put(a,1);
                      mape.put(a,b);
                    dp++;
                    bianli(a,tickets,h,map,n,dp,mape);
                }
                
                }
                
            }
        
    } 
    static void start(Map<String,Integer> map,String[][] tickets,int n){
        for(int i = 0;i<n;i++){
            map.put(tickets[i][0],0);
        }
    }
    public List<String> findItinerary(String[][] tickets) {
        int n = tickets.length;
        String a,b;
        int dp = 0;
        a="JFK";
        List<String> H= new ArrayList();
        H.add(a);
        Map map =new HashMap();
        start(map,tickets,n);
       Map mape =new HashMap();
        try {
             bianli(a,tickets,H,map,n,dp,mape);
        } catch (StopMsgException e) {
            return H;
        }
        return H;
    }
 
    static class StopMsgException extends RuntimeException {
    }
}