332. Reconstruct Itinerary

Description

You are given a list of airline tickets where tickets[i] = [fromi, toi] represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.

All of the tickets belong to a man who departs from "JFK", thus, the itinerary must begin with "JFK". 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"].

You may assume all tickets form at least one valid itinerary. You must use all the tickets once and only once.

Example 1:

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


Example 2:

Input: tickets = [["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.


Constraints:

• 1 <= tickets.length <= 300
• tickets[i].length == 2
• fromi.length == 3
• toi.length == 3
• fromi and toi consist of uppercase English letters.
• fromi != toi

Solutions

• class Solution {
void dfs(Map<String, Queue<String>> adjLists, List<String> ans, String curr) {
if (neighbors == null) {
return;
}
while (!neighbors.isEmpty()) {
String neighbor = neighbors.poll();
}
return;
}

public List<String> findItinerary(List<List<String>> tickets) {
Map<String, Queue<String>> adjLists = new HashMap<>();
for (List<String> ticket : tickets) {
String from = ticket.get(0);
String to = ticket.get(1);
}
}
List<String> ans = new ArrayList<>();
Collections.reverse(ans);
return ans;
}
}

• class Solution {
public:
vector<string> findItinerary(vector<vector<string>>& tickets) {
unordered_map<string, priority_queue<string, vector<string>, greater<string>>> g;
vector<string> ret;

// Initialize the graph
for (const auto& t : tickets) {
g[t[0]].push(t[1]);
}

findItineraryInner(g, ret, "JFK");

ret = {ret.rbegin(), ret.rend()};

return ret;
}

void findItineraryInner(unordered_map<string, priority_queue<string, vector<string>, greater<string>>>& g, vector<string>& ret, string cur) {
if (g.count(cur) == 0) {
// This is the end point
ret.push_back(cur);
return;
} else {
while (!g[cur].empty()) {
auto front = g[cur].top();
g[cur].pop();
findItineraryInner(g, ret, front);
}
ret.push_back(cur);
}
}
};

• class Solution:
def findItinerary(self, tickets: List[List[str]]) -> List[str]:
graph = defaultdict(list)

for src, dst in sorted(tickets, reverse=True):
graph[src].append(dst)

itinerary = []

def dfs(airport):
while graph[airport]:
dfs(graph[airport].pop())
itinerary.append(airport)

dfs("JFK")

return itinerary[::-1]


• func findItinerary(tickets [][]string) (ans []string) {
sort.Slice(tickets, func(i, j int) bool {
return tickets[i][0] > tickets[j][0] || (tickets[i][0] == tickets[j][0] && tickets[i][1] > tickets[j][1])
})
g := make(map[string][]string)
for _, ticket := range tickets {
g[ticket[0]] = append(g[ticket[0]], ticket[1])
}
var dfs func(f string)
dfs = func(f string) {
for len(g[f]) > 0 {
t := g[f][len(g[f])-1]
g[f] = g[f][:len(g[f])-1]
dfs(t)
}
ans = append(ans, f)
}
dfs("JFK")
for i := 0; i < len(ans)/2; i++ {
ans[i], ans[len(ans)-1-i] = ans[len(ans)-1-i], ans[i]
}
return
}

• function findItinerary(tickets: string[][]): string[] {
const g: Record<string, string[]> = {};
tickets.sort((a, b) => b[1].localeCompare(a[1]));
for (const [f, t] of tickets) {
g[f] = g[f] || [];
g[f].push(t);
}
const ans: string[] = [];
const dfs = (f: string) => {
while (g[f] && g[f].length) {
const t = g[f].pop()!;
dfs(t);
}
ans.push(f);
};
dfs('JFK');
return ans.reverse();
}