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leetcode题目：Clone Graph

2024-07-03 20:26:29   233人阅读

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 #.

  1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
  2. Second node is labeled as 1. Connect node 1 to node 2.
  3. 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
         /          \_/

思路：

1）遍历

2）用一张map存取旧指针和新指针，保持对应关系。

代码：

/**
 * Definition for undirected graph.
 * struct UndirectedGraphNode {
 *     int label;
 *     vector<UndirectedGraphNode *> neighbors;
 *     UndirectedGraphNode(int x) : label(x) {};
 * };
 */
class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if(node == NULL)
		{
			return NULL;
		}
		UndirectedGraphNode* newnode = new UndirectedGraphNode(node->label);
		m_map.insert(make_pair(node,newnode));
		checknewnode(node,newnode->neighbors);
		return newnode;
	}
private:
	map<UndirectedGraphNode*,UndirectedGraphNode*>m_map;
    //判断这条顶点的邻居是否是新邻居
	void checknewnode(UndirectedGraphNode* node,vector<UndirectedGraphNode*>&iivec)
	{
		vector<UndirectedGraphNode*>::iterator pos;
		for (pos=node->neighbors.begin();pos!=node->neighbors.end();pos++)
		{
			map<UndirectedGraphNode*,UndirectedGraphNode*>::iterator tmp = m_map.find(*pos);
			if (tmp==m_map.end())
			{
				UndirectedGraphNode* newnode = new UndirectedGraphNode((*pos)->label);
				m_map.insert(make_pair(*pos,newnode));
				checknewnode(*pos,newnode->neighbors);
				iivec.push_back(newnode);
			}
			else
			{
				iivec.push_back(tmp->second);
			}
		}
		
	}
};


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