Given an directed graph, a topological order of the graph nodes is defined as follow:
- For each directed edge
A -> Bin graph, A must before B in the order list. - The first node in the order can be any node in the graph with no nodes direct to it.
Find any topological order for the given graph.
Have you met this question in a real interview?
Yes
Example
For graph as follow:
The topological order can be:
[0, 1, 2, 3, 4, 5]
[0, 2, 3, 1, 5, 4]
...
Note
You can assume that there is at least one topological order in the graph.
Challenge
Can you do it in both BFS and DFS?
topSort:
/**
* 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
* @return: Any topological order for the given graph.
*/
public ArrayList<DirectedGraphNode> topSort(ArrayList<DirectedGraphNode> graph) {
// write your code here
ArrayList<DirectedGraphNode> res = new ArrayList<DirectedGraphNode>();
HashMap<DirectedGraphNode, Integer> map = new HashMap<DirectedGraphNode, Integer>();
for(DirectedGraphNode node : graph){
for(DirectedGraphNode neighbor : node.neighbors){
if(map.containsKey(neighbor)){
map.put(neighbor, map.get(neighbor) + 1);
} else {
map.put(neighbor, 1);
}
}
}
Queue<DirectedGraphNode> queue = new LinkedList<DirectedGraphNode>();
for(DirectedGraphNode node : graph){
if(!map.containsKey(node)){
queue.offer(node);
res.add(node);
}
}
while(!queue.isEmpty()){
DirectedGraphNode cur = queue.poll();
for(DirectedGraphNode neighbor : cur.neighbors){
map.put(neighbor, map.get(neighbor) - 1);
if(map.get(neighbor) == 0){
res.add(neighbor);
queue.offer(neighbor);
}
}
}
return res;
}
}
DFS: iterative
public class Solution {
/**
* @param graph: A list of Directed graph node
* @return: Any topological order for the given graph.
*/
public ArrayList<DirectedGraphNode> topSort(ArrayList<DirectedGraphNode> graph) {
// write your code here
ArrayList<DirectedGraphNode> res = new ArrayList<DirectedGraphNode>();
HashMap<DirectedGraphNode, Integer> map = new HashMap<DirectedGraphNode, Integer>();
for(DirectedGraphNode node : graph){
for(DirectedGraphNode neighbor : node.neighbors){
if(map.containsKey(neighbor)){
map.put(neighbor, map.get(neighbor) + 1);
} else {
map.put(neighbor, 1);
}
}
}
Stack<DirectedGraphNode> stack = new Stack<DirectedGraphNode>();
for(DirectedGraphNode node : graph){
if(!map.containsKey(node)){
stack.push(node);
res.add(node);
}
}
while(!stack.isEmpty()){
DirectedGraphNode cur = stack.pop();
for(DirectedGraphNode neighbor : cur.neighbors){
map.put(neighbor, map.get(neighbor) - 1);
if(map.get(neighbor) == 0){
res.add(neighbor);
stack.push(neighbor);
}
}
}
return res;
}
}
Recursive DFS:
public class Solution {
/**
* @param graph: A list of Directed graph node
* @return: Any topological order for the given graph.
*/
public ArrayList<DirectedGraphNode> topSort(ArrayList<DirectedGraphNode> graph) {
// write your code here
ArrayList<DirectedGraphNode> res = new ArrayList<DirectedGraphNode>();
Set<DirectedGraphNode> visited = new HashSet<DirectedGraphNode>();
for(DirectedGraphNode node : graph){
if(!visited.contains(node)){
dfs(visited, node, res);
}
}
return res;
}
private void dfs(Set<DirectedGraphNode> visited,DirectedGraphNode node, ArrayList<DirectedGraphNode> res){
visited.add(node);
for(DirectedGraphNode n : node.neighbors){
if(!visited.contains(n)){
dfs(visited, n, res);
}
}
res.add(0, node);
}
}
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