332. Reconstruct Itinerary

• 題目
  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.
   
  

• 解題思路

  • 本題是關於圖的邊進行遍歷，每張機票都是圖的一條有向邊，需要找出經過每條邊的路徑，並且必定有解本題，則對於某個節點（非起點）其只於一個節點相鄰且只存在一條邊，則這個節點必定是最後訪問的，否則不可能遍歷完所有邊，並且這種點最多一個（不包含起點）。

  • 解法 1 – DFS + 遞迴

    • 解決步驟
      • 將圖建立起來，建立鄰接表，使用map<string, multiset<string> 來儲存鄰接表。使用multiset可以自動排序。（set的預設排序由小到大，multiset預設排序是由大到小
        ）
      • 從節點JKF開始DFS遍歷，只要當前的對映集合multiset裡面還有節點，則取出這個節點，遞迴遍歷這個節點，同時需要將這個節點從multiset中刪除掉，當對映集合multiset為空的時候，則將節點加入到結果中
      • 因為當前儲存結果是回溯得到的，需要將結果的儲存順序反轉輸出
    • 實現程式碼

    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

    • 思路與解法一相同，利用資料結構stack進行迭代。
    • 實現程式碼

    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;
        }
    };