Leetcode 332 - Reconstruct itinerary

Note:

  • This is DFS but can be treated like backtracking (The hardest I’ve done so far)
  • We need to sort tickets first, so the first path we found must be the res coz we are using greedy algo to iterate in smallest lexical order.
  • We need a flag found because if we don’t stop the following forloop, more paths will be added into res.
  • Used visited to mark visited path on the same branch.

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

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Input: tickets = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
Output: ["JFK","MUC","LHR","SFO","SJC"]
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/**
* @param {string[][]} tickets
* @return {string[]}
*/
var findItinerary = function(tickets) {
let path = ['JFK'];
let ans = [];
let found = false;
let visited = [...Array(tickets.length).fill(false)];
tickets.sort((a,b) => {
if (a < b) return -1;
return 1;
})
backtracking("JFK");
return ans;

function backtracking(departure) {
if (path.length === tickets.length + 1) {
ans.push(...path);
found = true;
return;
}
for (let i = 0; i < tickets.length && !found; i++) {
if (tickets[i][0] !== departure || visited[i]) continue;
visited[i] = true;
path.push(tickets[i][1]);
backtracking(path[path.length-1]);
visited[i] = false;
path.pop();
}
}
};