Given a directed graph, design an algorithm to find out whether there is a route between two nodes.
Have you met this question in a real interview?
Yes
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
/**
* Definition for Directed graph.
* class DirectedGraphNode {
* int label;
* ArrayList<DirectedGraphNode> neighbors;
* DirectedGraphNode(int x) {
* label = x;
* neighbors = new ArrayList<DirectedGraphNode>();
* }
* };
*/
public class Solution {
/**
* @param graph: A list of Directed graph node
* @param s: the starting Directed graph node
* @param t: the terminal Directed graph node
* @return: a boolean value
*/
public boolean hasRoute(ArrayList<DirectedGraphNode> graph,
DirectedGraphNode s, DirectedGraphNode t) {
// write your code here
Set<DirectedGraphNode> visited = new HashSet<DirectedGraphNode>();
return dfs(s, t, visited);
}
private boolean dfs(DirectedGraphNode s, DirectedGraphNode t, Set<DirectedGraphNode> visited){
if(s == null) return false;
if(s.equals(t)) return true;
visited.add(s);
for(DirectedGraphNode n : s.neighbors){
if(visited.contains(n)) continue;
if(dfs(n, t, visited)) return true;
}
return false;
}
}
/**
* Definition for Directed graph.
* class DirectedGraphNode {
* int label;
* ArrayList<DirectedGraphNode> neighbors;
* DirectedGraphNode(int x) {
* label = x;
* neighbors = new ArrayList<DirectedGraphNode>();
* }
* };
*/
public class Solution {
/**
* @param graph: A list of Directed graph node
* @param s: the starting Directed graph node
* @param t: the terminal Directed graph node
* @return: a boolean value
*/
public boolean hasRoute(ArrayList<DirectedGraphNode> graph,
DirectedGraphNode s, DirectedGraphNode t) {
// write your code here
Set<DirectedGraphNode> visited = new HashSet<DirectedGraphNode>();
Stack<DirectedGraphNode> stack = new Stack<DirectedGraphNode>();
stack.push(s);
while(!stack.isEmpty()){
DirectedGraphNode node = stack.pop();
if(node.equals(t)) return true;
if(visited.contains(node)) continue;
for(DirectedGraphNode n : node.neighbors){
stack.push(n);
}
}
return false;
}
}
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