Problem

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
         / \
         \_/

思路

• need a hash table to build the relationship between original nodes and the copy.
• bfs to traverse the graph, need a queue
  • while the queue is not empty
    • poll the front node from the queue
    • loop through its neighbors to see whether the node has its copy and the node is visited.
    • no copy also means not being visited, then create copy and put to the map
    • no matter adj is visted, build the relationship between current copy node and its adjacent copy node
      • because relationship is bi-directional.

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

        UndirectedGraphNode ans = new UndirectedGraphNode(node.label);
        Map<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<>();
        map.put(node, ans);
        Queue<UndirectedGraphNode> queue = new LinkedList<>();
        queue.offer(node);
        while (!queue.isEmpty()) {
            UndirectedGraphNode curt = queue.poll();
            for (UndirectedGraphNode adj: curt.neighbors) {
                if (!map.containsKey(adj)) {
                    map.put(adj, new UndirectedGraphNode(adj.label));
                    queue.offer(adj);
                }
                map.get(curt).neighbors.add(map.get(adj));
            }
        }
        return ans;
    }