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，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
     
       neighbors; *     UndirectedGraphNode(int x) : label(x) {}; * }; */class Solution {public:    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {        unordered_map
      
        umap;        return clone(node, umap);    }    UndirectedGraphNode *clone(UndirectedGraphNode *node, unordered_map
       
         &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;    } };