Lintcode: Route Between Two Nodes in Graph

Given a directed graph, design an algorithm to find out whether there is a route between two nodes.

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Example
Given graph:

A----->B----->C
      |
      |
      |
      v
     ->D----->E
for s = B and t = E, return true

for s = D and t = C, return false

DFS:

 1 /**
 2  * Definition for Directed graph.
 3  * class DirectedGraphNode {
 4  *     int label;
 5  *     ArrayList<DirectedGraphNode> neighbors;
 6  *     DirectedGraphNode(int x) {
 7  *         label = x;
 8  *         neighbors = new ArrayList<DirectedGraphNode>();
 9  *     }
10  * };
11  */
12 public class Solution {
13    /**
14      * @param graph: A list of Directed graph node
15      * @param s: the starting Directed graph node
16      * @param t: the terminal Directed graph node
17      * @return: a boolean value
18      */
19     public boolean hasRoute(ArrayList<DirectedGraphNode> graph, 
20                             DirectedGraphNode s, DirectedGraphNode t) {
21         // write your code here
22         HashSet<DirectedGraphNode> visited = new HashSet<DirectedGraphNode>();
       visited.add(s);
23 return dfs(s, t, visited); 24 } 25 26 public boolean dfs(DirectedGraphNode s, DirectedGraphNode t, HashSet<DirectedGraphNode> visited) { 27 if (s == t) return true; 28 for (DirectedGraphNode neighbor : s.neighbors) { 29 if (!visited.contains(neighbor)) { 30 visited.add(s); 31 if (dfs(neighbor, t, visited)) 32 return true; 33 } 34 } 35 return false; 36 } 37 }

BFS:

 1 public class Solution {
 2    /**
 3      * @param graph: A list of Directed graph node
 4      * @param s: the starting Directed graph node
 5      * @param t: the terminal Directed graph node
 6      * @return: a boolean value
 7      */
 8     public boolean hasRoute(ArrayList<DirectedGraphNode> graph, 
 9                             DirectedGraphNode s, DirectedGraphNode t) {
10         // write your code here
11         HashSet<DirectedGraphNode> visited = new HashSet<DirectedGraphNode>();
12         LinkedList<DirectedGraphNode> queue = new LinkedList<DirectedGraphNode>();
13         if (s == t) return true;
14         queue.offer(s);
15         visited.add(s);
16         while (!queue.isEmpty()) {
17             DirectedGraphNode cur = queue.poll();
18             for (DirectedGraphNode neighbor : cur.neighbors) {
19                 if (neighbor == t) return true;
20                 if (visited.contains(neighbor)) continue;
21                 visited.add(neighbor);
22                 queue.offer(neighbor);
23             }
24         }
25         return false;
26     }
27 }
原文地址:https://www.cnblogs.com/EdwardLiu/p/5104283.html