Clone Graph

By | July 22, 2016
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leetcode 133.

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. Learned from here. Below is the DFS solution. I like this solution because it is a universal method. It can solve problem like Copy List with Random Pointer(github). Below is the code

public static UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
    return clone(node, new HashMap<>());
}

private static UndirectedGraphNode clone(UndirectedGraphNode node, HashMap<Integer, UndirectedGraphNode> hm) {
    if (node == null) {
        return null;
    }
    if (hm.containsKey(node.label)) {
        return hm.get(node.label);
    }
    UndirectedGraphNode cloneNode = new UndirectedGraphNode(node.label);
    hm.put(cloneNode.label, cloneNode);
    for (UndirectedGraphNode neighbor : node.neighbors) {
        cloneNode.neighbors.add(clone(neighbor, hm));
    }
    return cloneNode;
}

Check my code on github.