leetcode Clone Graph

Clone Graph

Total Accepted: 16482 Total
Submissions: 72324My Submissions

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
/ \
\_/

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//首先，这种深拷贝的题目，需要有一个辅助工具来判断中的节点是否重复。本题是图的拷贝，按照某一个规律全部把图的节点全部遍历一遍。

//我是用map来判断是否重复，然后根据原图进行广度优先遍历，然后把新图的节点一一new出来，并对节点进行相应的连接。

/**
* Definition for undirected graph.
* struct UndirectedGraphNode {
*     int label;
*     vector<UndirectedGraphNode *> neighbors;
*     UndirectedGraphNode(int x) : label(x) {};
* };
*/
//7:35->
class Solution {
public:
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
map<UndirectedGraphNode *,UndirectedGraphNode *> repeat;
// one set and two queue
// store the original nodes and new nodes;que is for old graph. que2 is for new graph.
queue<UndirectedGraphNode *> que;
queue<UndirectedGraphNode *> que2;
UndirectedGraphNode *head=NULL,*h1=NULL,*h2=NULL,*h3=NULL;
int i=0,j=0;
if(node==NULL)
{
return head;
}
que.push(node);
h1=head = new UndirectedGraphNode(node->label);
que2.push(h1);
repeat[node] = h1;
// begin bfs
while(que.size()>0)
{
node = que.front();
que.pop();
h1 = que2.front();
que2.pop();
for(i=0;i<node->neighbors.size();i++)
{
h2 = node->neighbors[i];
if(repeat.count(h2)==0)
{
h3 = new UndirectedGraphNode(h2->label);
h1->neighbors.push_back(h3);
repeat[h2] = h3;
que.push(h2);
que2.push(h3);
}else
{
//add the new link in new graph
h1->neighbors.push_back(repeat[h2]);
}
}
}
return head;
}
};