leetcode Clone Graph

题目连接

https://leetcode.com/problems/clone-graph/  

Clone Graph

Description

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 #.

First node is labeled as 0. Connect node 0 to both nodes 1 and 2. 
Second node is labeled as 1. Connect node 1 to node 2. 
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 {
private:
    typedef UndirectedGraphNode Node;
public:
    Node *cloneGraph(Node* node) {
        if(!node) return NULL;
        queue<Node*> q; q.push(node);
        unordered_map<Node*, Node*> vis;
        Node* head = new Node(node->label);
        vis[node] = head;
        while(!q.empty()) {
            Node* cur = q.front(); q.pop();
            vector<Node*> ret = cur->neighbors;
            for(auto &r: ret) {
                if(vis.find(r) == vis.end()) {
                    Node* tmp = new Node(r->label);
                    vis[cur]->neighbors.push_back(tmp);
                    vis[r] = tmp;
                    q.push(r);
                } else {
                    vis[cur]->neighbors.push_back(vis[r]);
                }
            }
        }
        return head;
    }
};