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 topSort(vector& graph) {
        // write your code here
        vector res;
        set visited;
    
        if(graph.size()==0)
            return graph;
        
        for(int i=0;ilabel) == visited.end())
                topSort(res, visited, graph[i]);
        }
        
        reverse(res.begin(), res.end());
        return res;
    }
    
private:
    void topSort(vector& res,
            set& 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);
    }

};

这种方法需要保证的是，第0个节点一定要是入度为零的节点，否则会出错。

第二种方法采用的是使用map统计所有节点的入度节点，然后使用queue保存入度为零的节点，最后使用bfs。

class Solution {
public:
    /**
     * @param graph: A list of Directed graph node
     * @return: Any topological order for the given graph.
     */
    vector topSort(vector graph) {
        // write your code here
        int size = graph.size(), i = 0;
        if(size <= 0) {
            return vector();
        }

        vector result;
        queue noPreNode;
        map nodeIndegree;

        getIndegree(graph, nodeIndegree);

        for(i=0; ineighbors.size(); i++) {
                if(--nodeIndegree[noPreNode.front()->neighbors[i]] == 0) {
                    noPreNode.push(noPreNode.front()->neighbors[i]);
                }
            }
            noPreNode.pop();
        }
        return result;
    }

    void getIndegree(vector &graph, map &nodeIndegree) {
        int size = graph.size();
        for(int i=0; ineighbors.size();
            for(int j=0; jneighbors[j]]++;
            }
        }
    }
};