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
         / \
         \_/

这道无向图的复制问题和之前的拷贝带有随机指针的链表有些类似，那道题的难点是如何处理每个节点的随机指针，这道题目的难点在于如何处理每个节点的neighbors，由于在深度拷贝每一个节点后，还要将其所有neighbors放到一个vector中，而如何避免重复拷贝呢？这道题好就好在所有节点值不同，所以我们可以使用哈希表来对应节点值和新生成的节点。对于图的遍历的两大基本方法是深度优先搜索DFS和广度优先搜索BFS，此题的两种解法可参见网友爱做饭的小莹子的博客，这里我们使用深度优先搜索DFS来解答此题，代码如下：

/**
 * Definition for undirected graph.
 * struct UndirectedGraphNode {
 *     int label;
 *     vector<UndirectedGraphNode *> neighbors;
 *     UndirectedGraphNode(int x) : label(x) {};
 * };
 */
class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        unordered_map<int, UndirectedGraphNode*> umap;
        return clone(node, umap);
    }
    UndirectedGraphNode *clone(UndirectedGraphNode *node, unordered_map<int, UndirectedGraphNode*> &umap) {
        if (!node) return node;
        if (umap.count(node->label)) return umap[node->label];
        UndirectedGraphNode *newNode = new UndirectedGraphNode(node->label);
        umap[node->label] = newNode;
        for (int i = 0; i < node->neighbors.size(); ++i) {
            (newNode->neighbors).push_back(clone(node->neighbors[i], umap));
        }
        return newNode;
    } 
};