LeetCode每日一题(332. Reconstruct Itinerary)

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

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我们根据 edges 可以得到每个 node 所需要到达的次数 stops， 然后从 JFK 出发，dfs 遍历整个 graph， 每走到一个 node, 我们将相应的 stops[node] -= 1, 在前往下一个 node 的时候我们将(source_node, destination_node)这个 edge 移出 edges。当 stops 中所有次数都为 0 时， 证明我们当前这条路径是可行的。

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use std::collections::HashMap;

impl Solution {
    fn dp(mut paths: HashMap<String, Vec<String>>, curr: &str, remain: &mut HashMap<String, i32>) -> Option<Vec<String>> {
        if let Some(n) = remain.remove(curr) {
            if n == 1 && remain.is_empty() {
                return Some(vec![curr.to_owned()]);
            }
            if n > 1 {
                remain.insert(curr.to_owned(), n - 1);
            }
            if let Some(mut next_places) = paths.remove(curr) {
                for i in 0..next_places.len() {
                    let place = next_places.remove(i);
                    let mut next_paths = paths.clone();
                    next_paths.insert(curr.to_owned(), next_places.clone());
                    if let Some(mut itinerary) = Solution::dp(next_paths, &place, remain) {
                        itinerary.insert(0, curr.to_owned());
                        return Some(itinerary);
                    }
                    next_places.insert(i, place.to_owned());
                    paths.insert(curr.to_owned(), next_places.clone());
                }
            }
            *remain.entry(curr.to_owned()).or_insert(0) += 1;
        }
        None
    }

    pub fn find_itinerary(tickets: Vec<Vec<String>>) -> Vec<String> {
        let mut remain = HashMap::new();
        remain.insert("JFK".into(), 1);
        let mut paths = HashMap::new();
        for t in tickets {
            *remain.entry(t[1].clone()).or_insert(0) += 1;
            paths.entry(t[0].clone()).or_insert(Vec::new()).push(t[1].clone());
        }
        for places in paths.values_mut() {
            places.sort();
        }
        Solution::dp(paths, "JFK", &mut remain).unwrap()
    }
}