Clone Graph

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
         / \
         \_/

/**
 * Definition for undirected graph.
 * struct UndirectedGraphNode {
 *     int label;
 *     vector<UndirectedGraphNode *> neighbors;
 *     UndirectedGraphNode(int x) : label(x) {};
 * };
 */
class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) 
    {
        typedef unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> MAPUNDIRECTED;
        queue<UndirectedGraphNode*> queueNode;
        MAPUNDIRECTED mapNode;
        if (0 == node)
            return 0;
        UndirectedGraphNode* newNode = new UndirectedGraphNode(node->label);
        mapNode[node] = newNode;
        queueNode.push(node);
        
        do
        {
            node = queueNode.front();
            queueNode.pop();
            MAPUNDIRECTED::iterator iter = mapNode.find(node);
            for (int i = 0; i < node->neighbors.size(); i++)
            {
                if (mapNode.end() == mapNode.find(node->neighbors[i]))
                {
                    UndirectedGraphNode* newNode = new UndirectedGraphNode(node->neighbors[i]->label);
                    pair<UndirectedGraphNode*, UndirectedGraphNode*> temp(node->neighbors[i], newNode);
                    mapNode.insert(temp);
                    queueNode.push(node->neighbors[i]);
                    iter->second->neighbors.push_back(newNode);
                }
                else
                    iter->second->neighbors.push_back(mapNode.find(iter->first->neighbors[i])->second);
            }
        } while (!queueNode.empty());
        return newNode;
    }
};