#127 Topological Sorting

题目描述：

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:

The topological order can be:

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

题目思路：

topological sort应该是经典算法之一吧，随便google两下就可以找到一堆解法。我是参考了下面的链接写的code：

DAG算法

Mycode（AC = 201ms）：

/** * Definition for Directed graph. * struct DirectedGraphNode { *     int label; *     vector<DirectedGraphNode *> neighbors; *     DirectedGraphNode(int x) : label(x) {}; * }; */class Solution {public:    /**     * @param graph: A list of Directed graph node     * @return: Any topological order for the given graph.     */    vector<DirectedGraphNode*> topSort(vector<DirectedGraphNode*> graph) {        // write your code here        vector<DirectedGraphNode*> ans;        unordered_set<int> visited;                for (int i = 0; i < graph.size(); i++) {            if (visited.find(graph[i]->label) == visited.end()) {                // for each unvisited node, find its reversed                // topsort                dfs(ans, visited, graph, graph[i]);            }        }                reverse(ans.begin(), ans.end());        return ans;    }        void dfs(vector<DirectedGraphNode*>& ans,             unordered_set<int>& visited,             vector<DirectedGraphNode*>& graph,             DirectedGraphNode* node)    {        if (visited.find(node->label) != visited.end()) {            return;        }                // use dfs to go finding the bottom unvisited node        for (int i = 0; i < node->neighbors.size(); i++) {            dfs(ans, visited, graph, node->neighbors[i]);        }                // once all the neighbors are found, then insert the current        // node        visited.insert(node->label);        ans.push_back(node);// 4, 1    }};