[leetcode] 133 Clone Graph

问题描述：

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
/ \
\_/

代码：

UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {  //c++
if(node == NULL)
return NULL;

set<int> visited;
map<int,UndirectedGraphNode*> myMap;
queue<UndirectedGraphNode*> myQueue;
UndirectedGraphNode *head,*p,*temp,*one;

myQueue.push(node);
bool isFirst = true;
do
{
p = myQueue.front();
if(isFirst){
temp = new UndirectedGraphNode(p->label);
head= temp;
myMap.insert(make_pair(p->label,temp));
isFirst = false;
}
else temp = myMap[p->label];
vector<UndirectedGraphNode*> vec;
for(int i=0; i<(p->neighbors).size(); i++){
int label =(p->neighbors)[i]->label;
if(myMap.count(label)!=0)
one = myMap[label];
else
{
one = new UndirectedGraphNode(label);
myMap.insert(make_pair(label,one));
}
vec.push_back(one);
if(visited.count(label)==0)
myQueue.push(p->neighbors[i]);
}
temp->neighbors = vec;
myQueue.pop();
visited.insert(p->label);
}while(!myQueue.empty());
return head;
}