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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

posted @ 2017-12-29 11:07  ranjiewen  阅读(134)  评论(0编辑  收藏  举报