Clone Graph

xioqua 发布于2019-08-14 17:07 / 3298人阅读

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 #.

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:

Solution

**S1. BFS get all nodes. (version with no level order)
S2. Use a Hashmap and store the mapping relationship btw old---new nodes.(create new node based on old one)
S3. get the neighbor info and connect the edge based on hashmap.**

 public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        if (node == null) {
            return node;
        }

        // use bfs algorithm to traverse the graph and get all nodes.
        ArrayList nodes = getNodes(node);
        
        // copy nodes, store the old->new mapping information in a hash map
        HashMap mapping = new HashMap<>();
        for (UndirectedGraphNode n : nodes) {
            mapping.put(n, new UndirectedGraphNode(n.label));
        }
        
        // copy neighbors(edges)
        for (UndirectedGraphNode n : nodes) {
            UndirectedGraphNode newNode = mapping.get(n);
            for (UndirectedGraphNode neighbor : n.neighbors) {
                UndirectedGraphNode newNeighbor = mapping.get(neighbor);
                newNode.neighbors.add(newNeighbor);
            }
        }
        
        return mapping.get(node);
    }
    
    private ArrayList getNodes(UndirectedGraphNode node) {
        Queue queue = new LinkedList();
        HashSet set = new HashSet<>();
        
        queue.offer(node);
        set.add(node);
        while (!queue.isEmpty()) {
            UndirectedGraphNode head = queue.poll();
            for (UndirectedGraphNode neighbor : head.neighbors) {
                if(!set.contains(neighbor)){
                    set.add(neighbor);
                    queue.offer(neighbor);
                }
            }
        }
        
        return new ArrayList(set);
    }

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