332. Reconstruct Itinerary

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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.

思路：刚开始用DFS，TLE，后来发现是求欧拉路径，如果不是要求大小顺序的话，不用heap，直接用个Set就好了

http://blog.csdn.net/stillxjy/article/details/51956183

https://discuss.leetcode.com/topic/36383/share-my-solution

import java.util.*;public class Solution {List<String> rst = new ArrayList<String>();Map<String, PriorityQueue<String>> neighbors = new HashMap<String, PriorityQueue<String>>();    public List<String> findItinerary(String[][] tickets) {        for(String[] ticket : tickets) {    if(!neighbors.keySet().contains(ticket[0]))    neighbors.put(ticket[0], new PriorityQueue<String>());    neighbors.get(ticket[0]).add(ticket[1]);    }        dfs("JFK");    return rst;    }        // Hierholzer's algorithmprivate void dfs(String src) {PriorityQueue<String> neighbor = neighbors.get(src);while(neighbor != null && !neighbor.isEmpty())dfs(neighbor.poll());rst.add(0, src);}}

感觉就是预先把能够到达的放到一个List里面，在dfs的时候就不需要每次遍历一遍判断

二刷：先遍历得到一条主路，但是不立马把结果放到rst里面，而是放到stack里面（因为可以会有环），然后主路遍历到头了，就开始从stack里面pop出来，当后退到有环路的路口点时，就不pop stack了，转而继续把值push到stack里。这样就实现了先把环里面的数先输出

import java.util.*;public class Solution {    public List<String> findItinerary(String[][] tickets) {                List<String> rst = new ArrayList<String>();    Map<String, PriorityQueue<String>> neighbors = new HashMap<String, PriorityQueue<String>>();        for(String[] ticket : tickets) {    if(!neighbors.keySet().contains(ticket[0]))    neighbors.put(ticket[0], new PriorityQueue<String>());    neighbors.get(ticket[0]).add(ticket[1]);    }        Stack<String> stack = new Stack<String>();    stack.add("JFK");    while(!stack.isEmpty()) {        String src = stack.peek();        if(neighbors.get(src)!=null && !neighbors.get(src).isEmpty()) {            stack.push(neighbors.get(src).remove());        } else {            rst.add(0, stack.pop());        }    }    return rst;    }}

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