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 #.

1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
2. Second node is labeled as 1. Connect node 1 to node 2.
3. 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);
            }
        }
    }
};