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  • 133. Clone Graph

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    133. Clone Graph

    题目

     Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.
    
    OJ's undirected graph serialization:
    
    Nodes are labeled uniquely.
    We use # as a separator for each node, and , as a separator for node label and each neighbor of the node.
    
    As an example, consider the serialized graph {0,1,2#1,2#2,2}.
    
    The graph has a total of three nodes, and therefore contains three parts as separated by #.
    
        First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
        Second node is labeled as 1. Connect node 1 to node 2.
        Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.
    
    Visually, the graph looks like the following:
    
           1
          / 
         /   
        0 --- 2
             / 
             \_/
    
    
    

    解析

    • 考察图的基本遍历方法,DFS/BFS
    • 注意细节bug
    • 运用unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> hash进行图的映射关系存储
    // clone graph
    class Solution_133 {
    
    // date 2017/12/29 10:01
    // date 2017/12/29 11:04
    public:
    
    	UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) { 
    		unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> hash;
    		if (!node)
    		{
    			return node;
    		}
    		if (hash.find(node)!=hash.end()) //找到,关键字已经访问过
    		{
    			hash[node] = new UndirectedGraphNode(node->label);
    			for (auto iter: node->neighbors)
    			{
    				hash[node]->neighbors.push_back(cloneGraph(iter)); //递归DFS   //超时
    			}
    		}
    
    		return hash[node];
    	}
    
    
    	UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) //BFS
    	{		
    		if (!node)
    		{
    			return node;
    		}	
    		unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> hash;
    		UndirectedGraphNode* head = new UndirectedGraphNode(node->label);
    		hash[node] = head;
    		queue<UndirectedGraphNode*> que;
    		que.push(node);  //que.push(head); bug 花费1小时查找
    
    		while (!que.empty())
    		{
    			UndirectedGraphNode* q = que.front();
    			que.pop();
    			for (auto iter: q->neighbors)
    			{
    				if (!hash[iter]) //还没有访问
    				{
    					UndirectedGraphNode* temp = new UndirectedGraphNode(iter->label);
    					hash[iter] = temp;
    					que.push(iter);
    				}
    
    				hash[q]->neighbors.push_back(hash[iter]); //将一个节点的邻接点关系记录下来
    			}
    			
    		}
    		return hash[node];
    	}
    
    };
    
    
    

    题目来源

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  • 原文地址:https://www.cnblogs.com/ranjiewen/p/8143454.html
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