LeetCode OJ - Clone Graph

题目：

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

解题思路：

　　递归求解，采用unoreded_map来记录新生成的节点。

代码：

/**
 * Definition for undirected graph.
 * struct UndirectedGraphNode {
 *     int label;
 *     vector<UndirectedGraphNode *> neighbors;
 *     UndirectedGraphNode(int x) : label(x) {};
 * };
 */
class Solution {
public:
    unordered_map<int, UndirectedGraphNode*> dict;
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if (node == NULL || dict.count(node->label)) return NULL;

        UndirectedGraphNode * new_node = new UndirectedGraphNode(node->label);
        dict[node->label] = new_node;

        for (int i = 0; i < (node->neighbors).size(); i++) {
            cloneGraph((node->neighbors)[i]);
            new_node->neighbors.push_back(dict[(node->neighbors)[i]->label]);
        }

        return new_node;
    }
};