L127 Topological Sort

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

• For each directed edgeA -> 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.

Notice

You can assume that there is at least one topological order in the graph.

Clarification

Learn more about representation of graphs

Example

For graph as follow:

picture

The topological order can be:

[0, 1, 2, 3, 4, 5]
[0, 2, 3, 1, 5, 4]
...

Challenge

Can you do it in both BFS and DFS?

拓扑排序...用bfs

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

    // collect in-degree
    HashMap<DirectedGraphNode, Integer> inDegree = new HashMap<>();
    for (DirectedGraphNode node : graph) {
        for (DirectedGraphNode nei : node.neighbors) {
            if (inDegree.containsKey(nei)) {
                inDegree.put(nei, inDegree.get(nei) + 1);
            } else {
                inDegree.put(nei, 1);
            }
        }
    }

    // put all indegree == 0 point in the result, also put in queue for tranvesal
    Queue<DirectedGraphNode> queue = new LinkedList<>();
    for (DirectedGraphNode node : graph) {
        if (!inDegree.containsKey(node)) {
            res.add(node);
            queue.offer(node);
        }
    }

    // do bfs and add ndoe to result
    while (!queue.isEmpty()) {
        DirectedGraphNode cur = queue.poll();

        for (DirectedGraphNode nei : cur.neighbors) {
            inDegree.put(nei, inDegree.get(nei) - 1);
            if (inDegree.get(nei) == 0) {
                inDegree.remove(nei);
                queue.offer(nei);
                res.add(nei);
            }
        }
    }

    if (inDegree.size() != 0) {
        // don't have topo order
    }

    return res;
}