#176 Route Between Two Nodes in Graph

最新推荐文章于 2021-05-08 17:30:11 发布

原创最新推荐文章于 2021-05-08 17:30:11 发布 · 397 阅读

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#lintcode #BFS

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本文介绍了一种使用广度优先搜索(BFS)算法来判断有向图中两个节点间是否存在路径的方法，并提供了一个具体的代码实现案例。

题目描述：

Given a directed graph, design an algorithm to find out whether there is a route between two nodes.

Have you met this question in a real interview?

Yes

Example

Given graph:

A----->B----->C
 \     |
  \    |
   \   |
    \  v
     ->D----->E

for s = B and t = E, return true

for s = D and t = C, return false

Tags

Cracking The Coding Interview Depth First Search Breadth First Search

题目思路：

这题我发现只能用bfs做，dfs做会超时（坑爹呀！）。bfs用简单的传统算法就好了。

Mycode（AC = 396ms）：

/**
 * 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
     * @param s: the starting Directed graph node
     * @param t: the terminal Directed graph node
     * @return: a boolean value
     */
    bool hasRoute(vector<DirectedGraphNode*> graph,
                  DirectedGraphNode* s, DirectedGraphNode* t) {
        // write your code here
        stack<DirectedGraphNode *> sgraph;
        sgraph.push(s);
        
        // bfs to search the neighbors
        while (!sgraph.empty()) {
            DirectedGraphNode *node = sgraph.top();
            sgraph.pop();
            
            if (node->label == t->label) return true;
            
            for (int i = 0; i < node->neighbors.size(); i++) {
                sgraph.push(node->neighbors[i]);
            }
        }
        
        return false;
    }
};