[LeetCode] 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
         / \
         \_/

思路：用map存储创建的新的结点，其中key是结点的label值，value是结点的指针值。

/**
 * Definition for undirected graph.
 * struct UndirectedGraphNode {
 *     int label;
 *     vector<UndirectedGraphNode *> neighbors;
 *     UndirectedGraphNode(int x) : label(x) {};
 * };
 */
class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if(node==NULL)
            return NULL;
        UndirectedGraphNode *pnode = node;
        UndirectedGraphNode *head = new UndirectedGraphNode(node->label);
        UndirectedGraphNode *p = head;
        map<int,UndirectedGraphNode *> labelAndNode;
        labelAndNode[head->label] = head;
        pNodeNeighbors(pnode,p,labelAndNode);
        return head;
    }
private:
    void pNodeNeighbors(UndirectedGraphNode *&pnode,UndirectedGraphNode *&p,map<int,UndirectedGraphNode *> &labelAndNode)
    {
        int num = pnode->neighbors.size();
        UndirectedGraphNode *pnew ;
        for(int i = 0;i<num;i++)
        {
            if(labelAndNode.count(pnode->neighbors[i]->label)==0)
            {
               pnew = new UndirectedGraphNode(pnode->neighbors[i]->label);
               labelAndNode[pnode->neighbors[i]->label] = pnew;
               p->neighbors.push_back(pnew);
               pNodeNeighbors(pnode->neighbors[i],p->neighbors[i],labelAndNode);
            }
            else
            {
              pnew =  labelAndNode[pnode->neighbors[i]->label];
              p->neighbors.push_back(pnew);
            }
        
        }
    
    
    }
};