## Clone Graph

06/28/2016 Graph Depth First Search Breadth First Search

## Question

Question 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 #.

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
/ \
\_/


## Solution

Result: Accepted Time: 78 ms

Here should be some explanations.

class Solution {
public:
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
unordered_map<int,UndirectedGraphNode *> mp;
queue<UndirectedGraphNode *> que;
UndirectedGraphNode * root = NULL;
if(node)
{
root = new UndirectedGraphNode(node->label);
que.push(node); que.push(root);
mp[root->label] = root;
}
while(!que.empty())
{
UndirectedGraphNode * old_node = que.front();que.pop();
UndirectedGraphNode * new_node = que.front();que.pop();
for(const auto & x:old_node->neighbors)
if(mp.count(x->label))
new_node->neighbors.push_back(mp[x->label]);
else
{
UndirectedGraphNode * tmp = new UndirectedGraphNode(x->label);
new_node->neighbors.push_back(tmp);
mp[x->label] = tmp;
que.push(x);que.push(tmp);
}
}
return root;
}
};


Complexity Analytics

• Time Complexity: $O(n)$
• Space Complexity: $O(1)$