Lintcode 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.

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：

这题就是topological sorting的算法之一：我使用dfs。

实现的是链接的算法--topological sorting

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*> res;
        set<int> visited;
    
        if(graph.size()==0)
            return graph;
        
        for(int i=0;i<graph.size();i++){
            if(visited.find(graph[i]->label) == visited.end())
                topSort(res, visited, graph[i]);
        }
        
        reverse(res.begin(), res.end());
        return res;
    }
    
private:
    void topSort(vector<DirectedGraphNode*>& res,
            set<int>& visited, DirectedGraphNode* node){
                
        if(visited.find(node->label) != visited.end())
            return;
           
        visited.insert(node->label);
        for(int i=0; i < node->neighbors.size(); i++){
            topSort(res, visited, node->neighbors[i]);
        }
        res.push_back(node);
    }

};