Reconstruct Itinerary
题目详情:
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.
Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.
解题方法:
使用dfs算法,从JFK开始遍历每个节点。要注意的是这里要求按照字典序访问节点,并且可能会有重复的节点即有两张相同的车票。所以这里使用map<string, multiset<string>>数据结构存储图是最简单的方式。
最开始我选择的方法是遍历到一个节点就将其加入到结果vector中,但是这样的缺点是当一个节点的子节点没有邻居时,不能先访问它,这样处理起来就有点麻烦。后来参考 discuss中的算法,按序访问节点的每一个邻居,并将其删除,当节点没有邻居时,再将其加入到结果vector中。最后反转结果vector。
代码详情:
class Solution {
public:
vector<string> findItinerary(vector<pair<string, string>> tickets) {
for (auto ticket : tickets)
targets[ticket.first].insert(ticket.second);
visit("JFK");
return vector<string>(route.rbegin(), route.rend());
}
map<string, multiset<string>> targets;
vector<string> route;
void visit(string airport) {
while (targets[airport].size()) {
string next = *targets[airport].begin();
targets[airport].erase(targets[airport].begin());
visit(next);
}
route.push_back(airport);
}
};
public:
vector<string> findItinerary(vector<pair<string, string>> tickets) {
for (auto ticket : tickets)
targets[ticket.first].insert(ticket.second);
visit("JFK");
return vector<string>(route.rbegin(), route.rend());
}
map<string, multiset<string>> targets;
vector<string> route;
void visit(string airport) {
while (targets[airport].size()) {
string next = *targets[airport].begin();
targets[airport].erase(targets[airport].begin());
visit(next);
}
route.push_back(airport);
}
};