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leetcode 332 Reconstruct Itinerary

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leetcode 332 Reconstruct Itinerary

原题链接:https://leetcode.com/problems/reconstruct-itinerary/


Description

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:
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.


Solution

题目大意: 从”JFK”出发找出一个序列使得所有的tickets用完,注意所求的结果要求字典序最小.
思路: 深搜,注意要标记已经访问过的节点(回溯时要注意节点还原)。
PS: 用完

n
张票当然需要用n+1个点



class Solution {
public:
vector findItinerary(vector> tickets) {
vector res;
if(tickets.empty()) return res;
for(auto &r: tickets) mp[r.first].insert({ r.second, false });
res.push_back("JFK");
dfs("JFK", res, tickets.size());
return res;
}
bool dfs(string from, vector &res, int n) {
if((int)res.size() == n + 1) return true;
for(auto r = mp[from].begin(); r != mp[from].end(); ++r) {
string temp = (*r).first;
auto c = mp[from].find({ temp, false });
if(c == mp[from].end()) continue;
res.push_back(temp);
mp[from].erase(c);
mp[from].insert({ temp, true });
if(dfs(temp, res, n)) return true;
res.pop_back();
c = mp[from].find({ temp, true });
if(c == mp[from].end()) continue;
mp[from].erase(c);
mp[from].insert({ temp, false });
}
return false;
}
private:
unordered_map>> mp;
};



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