回溯法解332. Reconstruct Itinerary
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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: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.
题意分析
题目意思是说,给定一系列的飞机出发地和飞机目的地,求从JFK出发的行程,这个行程的结果要求是最小序列(如果有多个可能行程序列的话)
思路分析
这道题跟course schedule的思路分析相似,course schedule的解法是针对点的dfs,这道题解法是针对边的dfs,在这里提供另外一种类似的思路,利用栈来进行回溯,因为题目要求是最小序列,因此可以使用multiset来存储这个图。开始遍历图中的边,只要图中的边没有遍历完,那么就把遍历到的点加进栈中,然后继续遍历,最后把栈的点依次加入到vector的首位,则得到最终结果
AC代码
class Solution {public: vector<string> findItinerary(vector<pair<string, string>> tickets) { vector<string> result; stack<string> backtrack; backtrack.push("JFK"); unordered_map<string, multiset<string>> mapstr; for (int i = 0; i < tickets.size(); ++i) { mapstr[tickets[i].first].insert(tickets[i].second); } while (!backtrack.empty()) { string currentdes = backtrack.top(); if (mapstr[currentdes].empty()) { result.insert(result.begin(), currentdes); backtrack.pop(); } else { backtrack.push(*mapstr[currentdes].begin()); mapstr[currentdes].erase(mapstr[currentdes].begin()); } } return result; }};
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