LeetCode:332. Reconstruct Itinerary(Week 5)

332. Reconstruct Itinerary

• 题目
  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:

  Input: [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
  Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]
  

  Example 2:

  Input: [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
  Output: ["JFK","ATL","JFK","SFO","ATL","SFO"]
  Explanation: Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"].But it is larger in lexical order.
  

• 解题思路

  • 本题是关于图的边进行遍历，每张机票都是图的一条有向边，需要找出经过每条边的路径，并且必定有解本题，则对于某个节点（非起点）其只于一个节点相邻且只存在一条边，则这个节点必定是最后访问的，否则不可能遍历完所有边，并且这种点最多一个（不包含起点）。

  • 解法 1 – DFS + 递归

    • 解决步骤
      • 将图建立起来，建立邻接表，使用map<string, multiset<string> 来存储邻接表。使用multiset可以自动排序。（set的默认排序由小到大，multiset默认排序是由大到小）
      • 从节点JKF开始DFS遍历，只要当前的映射集合multiset里面还有节点，则取出这个节点，递归遍历这个节点，同时需要将这个节点从multiset中删除掉，当映射集合multiset为空的时候，则将节点加入到结果中
      • 因为当前存储结果是回溯得到的，需要将结果的存储顺序反转输出
    • 实现代码

    class Solution {
    public:
        vector<string> findItinerary(vector<pair<string, string>> tickets) {
            vector<string> v;
            map<string, multiset<string> > myMap;
            for(auto it : tickets)
            	myMap[it.first].insert(it.second);
    
            dfs("JFK", v, myMap);
            reverse(v.begin(), v.end());
            return v;
        }
    
        void dfs(string start, vector<string>& v, map<string, multiset<string> > &myMap) {
        	while(myMap[start].size() > 0) {
        		string next = *myMap[start].begin();
        		myMap[start].erase(myMap[start].begin());
        		dfs(next, v, myMap);
        	}
        	v.push_back(start);
        }
    };
    

  • 解法 2 – DFS + 迭代

    • 思路与解法一相同，利用数据结构stack进行迭代。
    • 实现代码

    class Solution {
    public:
        vector<string> findItinerary(vector<pair<string, string>> tickets) {
            vector<string> v;
            map<string, multiset<string> > myMap;
            for(auto it : tickets)
            	myMap[it.first].insert(it.second);
    
            stack<string> myStack;
            myStack.push("JFK");
            while(!myStack.empty()) {
            	string node = myStack.top();
            	
            	if(!myMap[node].size()) {
            		myStack.pop();
            		v.push_back(node);
            	} 
            	else {
            		myStack.push(*myMap[node].begin());
            		myMap[node].erase(myMap[node].begin());
            	}
            }
            reverse(v.begin(), v.end());
            return v;
        }
    };