LintCode Q 127 - Topological Sorting

Given an directed graph, a topological order of the graph nodes is defined as follow:

For each directed edge A -> B in 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.

Challenge: Can you do it in both BFS and DFS?

Solution

Solution 1 : DFS

Definition for Directed graph.

class DirectedGraphNode {
	int label;
	ArrayList<DirectedGraphNode> neighbors;
	DirectedGraphNode(int x) { label = x; neighbors = new ArrayList<DirectedGraphNode>(); }
};

Code:

public ArrayList<DirectedGraphNode> topSort(ArrayList<DirectedGraphNode> graph) {
	ArrayList<DirectedGraphNode> res = new ArrayList<>();
	if (graph == null || graph.size() == 0) return res;

	Set<DirectedGraphNode> visited = new HashSet<>();

	for (DirectedGraphNode node: graph) 
		dfs(node, visited, res);

	return res;
}

private void dfs(DirectedGraphNode node, Set<DirectedGraphNode> visited, ArrayList<DirectedGraphNode> res) {

	if (visited.contains(node)) return;

	visited.add(node);
	for (DirectedGraphNode next: node.neighbors) {
		dfs(next, visited, res);
	}

	res.add(0, node);

	return;
}

Solution 2 : BFS / Topology Sorting

Code:

public ArrayList<DirectedGraphNode> topSort(ArrayList<DirectedGraphNode> graph) {
	ArrayList<DirectedGraphNode> res = new ArrayList<>();
	if (graph == null || graph.size() == 0) return res;

	int n = graph.size();
	int[] order = new int[n];
	for (DirectedGraphNode node: graph) {
		for (DirectedGraphNode next: node.neighbors) {
			order[next.label]++;
		}
	}

	Queue<DirectedGraphNode> que = new LinkedList<>();
	for (DirectedGraphNode node: graph) {
		if (order[node.label] == 0) que.offer(node);
	}

	while (!que.isEmpty()) {
		DirectedGraphNode curr = que.poll();
		res.add(curr);
		for (DirectedGraphNode next: curr.neighbors) {
			order[next.label]--;
			if (order[next.label] == 0) que.offer(next);
		}
	}

	return res;
}

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