LeetCode133：Clone Graph

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

#

.

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
/ \
\_/

解题思路：

本题要解决的问题就是对一个图进行clone，这不禁让我们想起图的遍历：DFS和BFS，所以我们可以再遍历图中某个点时，增加一些附加的操作而不仅仅是访问它，这里我们要增加的附加操作即对该点进行clone。

现在我们利用DFS和BFS两种方式进行解题。

实现代码：

#include <iostream>
#include <vector>
#include <queue>
#include <unordered_map>
using namespace std;

/*
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
/ \
\_/

*/

struct UndirectedGraphNode {
int label;
vector<UndirectedGraphNode *> neighbors;
UndirectedGraphNode(int x) : label(x) {};
};

class Solution {
public:
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
if(node == NULL)
return NULL;
unordered_map<int, UndirectedGraphNode*> hashtable;//标志位，是否已拷贝了key为label，value为新拷贝的node
return dfsclone(node, hashtable);

}

//DFS
UndirectedGraphNode *dfsclone(UndirectedGraphNode *node, unordered_map<int, UndirectedGraphNode*> &hashtable)
{
if(node == NULL)
return NULL;
if(hashtable.count(node->label) > 0)
return hashtable[node->label];//已经拷贝好了，则直接返回即可，否则进行下面的拷贝操作
UndirectedGraphNode *copyNode = new UndirectedGraphNode(node->label);
hashtable[copyNode->label] = copyNode;
vector<UndirectedGraphNode *>::iterator iter;
for(iter = node->neighbors.begin(); iter != node->neighbors.end(); ++iter)
{
copyNode->neighbors.push_back(dfsclone(*iter, hashtable));//进行深度优先算法拷贝
}

return copyNode;
}

//BFS
UndirectedGraphNode *bfsclone(UndirectedGraphNode *node, unordered_map<int, UndirectedGraphNode*> &hashtable)
{
if(node == NULL)
return NULL;
queue<UndirectedGraphNode *> qu;
UndirectedGraphNode *copyNode = new UndirectedGraphNode(node->label);
hashtable[copyNode->label] = copyNode;
qu.push(node);
while(!qu.empty())
{
UndirectedGraphNode *tnode = qu.front();
qu.pop();
vector<UndirectedGraphNode *>::iterator iter;
for(iter = node->neighbors.begin(); iter != node->neighbors.end(); ++iter)
{
if(hashtable.count((*iter)->label) == 0)
{
UndirectedGraphNode *tnode = new UndirectedGraphNode((*iter)->label);
hashtable[(*iter)->label] = tnode;
qu.push(*iter);
}
(hashtable[tnode->label])->neighbors.push_back(hashtable[(*iter)->label]);

}

}
return copyNode;
}
};
int main(void)
{
return 0;
}