LeetCode（133） Clone Graph

题目：复制一张无向图.

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
/ \
\_/

解法：
遍历一遍原图即可，可以用bfs或dfs。容易错的是图有可能有多重边。另外邻接表中的元素是指针，所以如果访问某个节点n，它的邻居中有一个已经生成过的节点，要找到原来的这个节点，再把指向原来节点的指针放入n的邻接表中。否则，生成一个新节点，再把指向新节点的指针放入n的邻接表中。
代码：
class Solution {
public:
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
if(node==NULL) return NULL;
map<int,UndirectedGraphNode*> hasCreated;//已生成的节点集合
UndirectedGraphNode *graph=new UndirectedGraphNode(node->label);
hasCreated[node->label]=graph;
dfs(graph,node,hasCreated);
return graph;
}
void dfs(UndirectedGraphNode *newGraph,UndirectedGraphNode *oldGraph,map<int,UndirectedGraphNode*>& hasCreated)
{
for(int i=0;i<(oldGraph->neighbors).size();i++)
{
UndirectedGraphNode *neighbor=(oldGraph->neighbors)[i];
map<int,UndirectedGraphNode*>::iterator it=hasCreated.find(neighbor->label);
if(it==hasCreated.end())
{
UndirectedGraphNode *node=new UndirectedGraphNode(neighbor->label);
(newGraph->neighbors).push_back(node);
hasCreated[neighbor->label]=node;
dfs(node,neighbor,hasCreated);
}
else
{
(newGraph->neighbors).push_back(it->second);
}
}
}
};