LeetCode 332. Reconstruct Itinerary（重构行程）

原题网址：https://leetcode.com/problems/reconstruct-itinerary/

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:

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"]

.
All airports are represented by three capital letters (IATA code).
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.
这是一道经典的图论问题，本质上是欧拉路径/七桥问题/一笔画问题。

方法一：深度优先搜索，如果无法遍历全部边（机票），则回溯。

public class Solution {
private boolean find(Map<String, List<String>> graph, List<String> itinerary, int target) {
if (itinerary.size() == target) return true;

String t = itinerary.get(itinerary.size()-1);
List<String> next = graph.get(t);
if (next == null) return false;
for(int i=0; i<next.size(); i++) {
itinerary.add(next.get(i));
next.remove(i);
if (find(graph, itinerary, target)) return true;
next.add(i, itinerary.get(itinerary.size()-1));
itinerary.remove(itinerary.size()-1);
}
return false;
}
public List<String> findItinerary(String[][] tickets) {
Map<String, List<String>> graph = new HashMap<>();
for(int i=0; i<tickets.length; i++) {
List<String> next = graph.get(tickets[i][0]);
if (next == null) {
next = new ArrayList<>();
graph.put(tickets[i][0], next);
// set.add(tickets[i][0]);
}
next.add(tickets[i][1]);
}
for(List<String> next: graph.values()) Collections.sort(next);
List<String> itinerary = new ArrayList<>();
itinerary.add("JFK");
find(graph, itinerary, tickets.length+1);
return itinerary;
}
}

方法二：直接应用一笔画算法。

public class Solution {
public List<String> findItinerary(String[][] tickets) {
Map<String, PriorityQueue<String>> graph = new HashMap<>();
for(String[] ticket: tickets) {
PriorityQueue<String> nexts = graph.get(ticket[0]);
if (nexts == null) {
nexts = new PriorityQueue<>();
graph.put(ticket[0], nexts);
}
nexts.add(ticket[1]);
}
return eulerian(graph, "JFK");
}

private List<String> eulerian(Map<String, PriorityQueue<String>> graph, String city) {
List<String> route = new ArrayList<>();
route.add(city);
PriorityQueue<String> nexts = graph.get(city);
if (nexts == null) return route;
while (!nexts.isEmpty()) {
String next = nexts.remove();
route.addAll(1, eulerian(graph, next));
}
return route;
}
}

方法三：一笔画算法的递归实现（参考LeetCode的论坛文章https://leetcode.com/discuss/84659/short-ruby-python-java-c）

public class Solution {
public List<String> findItinerary(String[][] tickets) {
Map<String, PriorityQueue<String>> graph = new HashMap<>();
List<String> route = new ArrayList();
for(String[] ticket: tickets) {
PriorityQueue<String> heap = graph.get(ticket[0]);
if (heap == null) {
heap = new PriorityQueue<>();
graph.put(ticket[0], heap);
}
heap.add(ticket[1]);
}
visit(graph, route, "JFK");
return route;
}

private void visit(Map<String, PriorityQueue<String>> graph, List<String> route, String airport) {
PriorityQueue<String> next = graph.get(airport);
while (next != null && !next.isEmpty()) visit(graph, route, next.poll());
route.add(0, airport);
}
}

另一种实现：

public class Solution {
private void find(String curr, Map<String, List<String>> graph, List<String> itinerary) {
List<String> nexts = graph.get(curr);
while (nexts != null && !nexts.isEmpty()) {
String next = nexts.get(0);
nexts.remove(0);
find(next, graph, itinerary);
}
itinerary.add(0, curr);
}
public List<String> findItinerary(String[][] tickets) {
Map<String, List<String>> graph = new HashMap<>();
for(String[] ticket: tickets) {
List<String> nexts = graph.get(ticket[0]);
if (nexts == null) {
nexts = new ArrayList<>();
graph.put(ticket[0], nexts);
}
nexts.add(ticket[1]);
}
for(List<String> nexts: graph.values()) Collections.sort(nexts);
List<String> itinerary = new ArrayList<>();
find("JFK", graph, itinerary);
return itinerary;
}
}