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:

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;
}