lintcode:Topological Sorting

 Topological SortingShow result 

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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.

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 Notice

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

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For graph as follow:

顺便贴一个topology的视频：https://www.youtube.com/watch?v=ddTC4Zovtbc

/** * Definition for Directed graph. * struct DirectedGraphNode { *     int label; *     vector<DirectedGraphNode *> neighbors; *     DirectedGraphNode(int x) : label(x) {}; * }; */class Solution {    private:    void topSortHelper(DirectedGraphNode* curNode, stack<DirectedGraphNode*> &auxStack, set<int> &visited)    {        visited.insert(curNode->label);                for (int i=0; i<curNode->neighbors.size(); i++)        {            if (visited.find(curNode->neighbors[i]->label) == visited.end())             {                topSortHelper(curNode->neighbors[i], auxStack, visited);            }        }        auxStack.push(curNode);//当这个节点已经visit 过它所有的子节点的之后，就把它发到stack中，所以stack的最上层肯定是依赖性最小的，食物链最高的节点    }        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                set<int> visited;        stack<DirectedGraphNode*> auxStack;                for (int i=0; i<graph.size(); i++)        {            if (visited.find(graph[i]->label) == visited.end())            {                topSortHelper(graph[i], auxStack, visited);            }        }                vector<DirectedGraphNode*> retVtr;        while (auxStack.size() > 0)        {            retVtr.push_back(auxStack.top());            auxStack.pop();        }                return retVtr;    }};