题目：
Clone an undirected graph. Each node in the graph contains a label and
a list of its neighbors.
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 #.
0.
Connect node 0 to both nodes 1 and 2.1.
Connect node 1 to node 2.2.
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);
}
}
}
};leetcode题目：Clone Graph,布布扣,bubuko.com
原文地址：http://blog.csdn.net/woailiyin/article/details/25398011
