【LeetCode】332. Reconstruct Itinerary 重新安排行程（Medium）（JAVA）

【LeetCode】332. Reconstruct Itinerary 重新安排行程（Medium）（JAVA）

题目地址: https://leetcode.com/problems/reconstruct-itinerary/

题目描述：

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.
4. One must use all the tickets once and only once.

Example 1:

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

Example 2:

Input: [["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.

题目大意

给定一个机票的字符串二维数组 [from, to]，子数组中的两个成员分别表示飞机出发和降落的机场地点，对该行程进行重新规划排序。所有这些机票都属于一个从 JFK（肯尼迪国际机场）出发的先生，所以该行程必须从 JFK 开始。

提示：

1. 如果存在多种有效的行程，请你按字符自然排序返回最小的行程组合。例如，行程 [“JFK”, “LGA”] 与 [“JFK”, “LGB”] 相比就更小，排序更靠前
2. 所有的机场都用三个大写字母表示（机场代码）。
3. 假定所有机票至少存在一种合理的行程。
4. 所有的机票必须都用一次 且 只能用一次。

解题方法

1. 用一个 map 来存对应飞往的地方，目的地用优先队列来存，让字符顺序小的在前面
2. 因为肯定存在一个解，所以优先遍历小的，然后遍历到结束，这样存在一个问题，可能小的这条路最终到了终点，但是有的没遍历过
3. 所以我们采用逆序的方式，如果遍历到了死胡同，他就优先加入，最后翻转，就在最后了；如果没有遍历到死胡同，那就是刚好按照顺序遍历完了，加入的时候也是后加入的，翻转就是结果

class Solution {
    public List<String> findItinerary(List<List<String>> tickets) {
        if (tickets.size() == 0) return new ArrayList<>();
        Map<String, PriorityQueue<String>> map = new HashMap<>();
        List<String> list = new ArrayList<>();
        for (int i = 0; i < tickets.size(); i++) {
            PriorityQueue<String> temp = map.get(tickets.get(i).get(0));
            if (temp == null) {
                temp = new PriorityQueue<>();
                map.put(tickets.get(i).get(0), temp);
            }
            temp.offer(tickets.get(i).get(1));
        }
        fH(map, list, "JFK");
        Collections.reverse(list);
        return list;
    }

    public void fH(Map<String, PriorityQueue<String>> map, List<String> list, String cur) {
        PriorityQueue<String> nextQueue = map.get(cur);
        while (nextQueue != null && nextQueue.size() > 0) {
            fH(map, list, nextQueue.poll());
        }
        list.add(cur);
    }
}

执行耗时:6 ms,击败了99.32% 的Java用户
内存消耗:39.2 MB,击败了77.96% 的Java用户

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