#127 Topological Sorting

最新推荐文章于 2024-06-05 21:23:28 发布

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最新推荐文章于 2024-06-05 21:23:28 发布

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文章标签： lintcode DFS topological sorting

本文链接：https://blog.youkuaiyun.com/haifischxia/article/details/52299659

本文介绍了一种基于深度优先搜索（DFS）实现的图的拓扑排序算法，并提供了具体的C++代码示例。该算法适用于有向无环图（DAG），能够找到至少一种合法的节点排序方式。

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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：

这题就是。。。topological sorting的算法之一：我用了dfs的code，实现的是链接的算法--topological sorting

Mycode (AC = 162ms):

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