[leetcode#133] Clone Graph

The problem:

Clone an undirected graph. Each node in the graph contains a and label a lists of its neighbors .

OJ ' s undirected graph serialization:

Nodes is 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 to 0 both nodes 1 and 2 .
2. Second node is labeled as 1 . Connect node to 1 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         /          \_/

My Analysis:

The implementation pattern used in Thisproblem is very classic in cloning a graph. The key Idea:use a hash map to store and retrieve the copy of a node. <original, copy>The basic idea was to use the invariant:1. We dequeue a node from the queue and then we scan the node ' s neighbors. We check if the neighbor node has already been traversaled by using the "Containkeys ()" method. 1.1 IFF the node is in the HashMap, we would don't make a copy forit.1.2 IFF the node is not in the hashmap and we make a copy for  ThisNeighbor node. Put the <neighbor_node, copy>Into the HashMap and enqueue the neighbor node.if(!map.containskey (Cur.neighbors.get (i))) {//This node have not been traversaledcopy =NewUndirectedgraphnode (Cur.neighbors.get (i). label);        Map.put (Cur.neighbors.get (i), copy); Queue.offer (Cur.neighbors.get (i));  Note1:we only enqueue nodes from neighbors list, which means all neighbors list of remaining nodes in the queue (until the dequeue) have not been copyed yet. Thus, forEach of the current node's Neghbor list, we need to fully copy it. for(inti = 0; I < cur.neighbors.size (); i++) {    if(!Map.containskey (Cur.neighbors.get (i)))    {    ...    } Map.get (cur). Neighbors.add (Map.get (Cur.neighbors.get (i))); Note2:to Use the invariant, we need-make necessary preparation forIt (dummy layer). We should suppose there is dummy layer copy of the starting node, and then put it into HashMap and enqueue the starting node.

My Solution:

 Public classSolution { Publicundirectedgraphnode Clonegraph (Undirectedgraphnode node) {if(node = =NULL)            return NULL; HashMap<undirectedgraphnode, undirectedgraphnode> map =NewHashmap<undirectedgraphnode, undirectedgraphnode> (); Queue<UndirectedGraphNode> queue =NewLinkedlist<undirectedgraphnode> ();                Undirectedgraphnode cur; Undirectedgraphnode Copy=NewUndirectedgraphnode (Node.label);        Queue.offer (node);                Map.put (node, copy);  while(!Queue.isempty ()) {cur=Queue.poll ();  for(inti = 0; I < cur.neighbors.size (); i++) {                if(!map.containskey (Cur.neighbors.get (i))) {//This node have not been traversaledcopy =NewUndirectedgraphnode (Cur.neighbors.get (i). label);                    Map.put (Cur.neighbors.get (i), copy);                Queue.offer (Cur.neighbors.get (i));            } map.get (cur). Neighbors.add (Map.get (Cur.neighbors.get (i))); }        }                returnmap.get (node); }}

[leetcode#133] Clone Graph