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;
}
};