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.

思考：
BFS (??)
使用queue

/**
 * 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
        if (graph == null || graph.size() == 0) {
            return graph;
        }
       
        ArrayList<DirectedGraphNode> rst = new ArrayList<>();
       
       
        // keep track of all neighbors in HashMap
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        // 找到所有被directed的点并记录被directed多少次
        for (DirectedGraphNode node : graph) {
            for (DirectedGraphNode neighbor : node.neighbors) {
                int key = neighbor.label;
                if (map.containsKey(key)) {
                    map.put(key, map.get(key) + 1);
                } else {
                    map.put(key, 1);
                }
            }
        }
       
        // find all the root which never be directed and cannot be found in map
        Queue<DirectedGraphNode> queue = new LinkedList<> ();
        for (DirectedGraphNode node : graph) {
            if (!map.containsKey(node.label)) {
                queue.offer(node);
                rst.add(node);
            }
        }
       
       
        // BFS: according the node in queue to proccess
        while (!queue.isEmpty()) {
            DirectedGraphNode node = queue.poll();
            for (DirectedGraphNode neighbor : node.neighbors) {
                int key = neighbor.label;
                map.put(key, map.get(key)-1);
                if(map.get(key) == 0) {
                    queue.offer(neighbor);
                    rst.add(neighbor);
                }
            }
        }
        return rst;
       
    }
   
}