description： 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.
Notice
You can assume that there is at least one topological order in the graph.
Have you met this question in a real interview? Yes 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] …
出現(xiàn)了一個(gè)問題，hashset是存入和取出是沒有規(guī)律的，但是這是有向圖的問題，因此因該使用arraylist來(lái)進(jìn)行記錄
/** * 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) { return null; } Map<DirectedGraphNode, Integer> map = new HashMap<>(); for (DirectedGraphNode node : graph) { for (DirectedGraphNode root : node.neighbors) { if (map.containsKey(root)) { map.put(root, map.get(root) + 1); } else { map.put(root, 1); } } } Queue<DirectedGraphNode> queue = new LinkedList<>(); ArrayList<DirectedGraphNode> set = new ArrayList<>(); for (DirectedGraphNode node : graph) { if(!map.containsKey(node)) { set.add(node); queue.offer(node); } } while (!queue.isEmpty()) { DirectedGraphNode root = queue.poll(); for (DirectedGraphNode node : root.neighbors) { map.put(node, map.get(node) - 1); if (map.get(node) == 0) { queue.offer(node); set.add(node); } } } return new ArrayList<DirectedGraphNode>(set); }}新聞熱點(diǎn)
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