Clone Graph——LeetCode

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
         / 
         \_/

题目大意是给定一个无向图,可能有环,实现一个方法,deep clone这个图。

我的做法是采用BFS,从一个点开始,遍历它的邻居,然后加入队列,当队列非空,循环遍历,因为label是唯一的,采用map保存新生成的node,用label作为key。另外使用visited数组保存是否加入过队列,以免重复遍历。

Talk is cheap>>

  public UndirectedGraphNode cloneGraph(UndirectedGraphNode root) {
        HashSet<Integer> visited = new HashSet<>();
        if (root==null)
            return null;
        List<UndirectedGraphNode> queue = new ArrayList<>();
        HashMap<Integer,UndirectedGraphNode> map = new HashMap<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            UndirectedGraphNode node = queue.get(0);
            if (map.get(node.label)==null) {
                map.put(node.label, new UndirectedGraphNode(node.label));
            }
            UndirectedGraphNode tmp = map.get(node.label);
            queue.remove(0);
         
            for (int i = 0; i < node.neighbors.size(); i++) {
                int key = node.neighbors.get(i).label;
                if (map.get(key)==null){
                    map.put(key,new UndirectedGraphNode(key));
                }
                tmp.neighbors.add(map.get(key));
                visited.add(node.label);
                if (!visited.contains(key)){
                    queue.add(node.neighbors.get(i));
                    visited.add(key);
                }
            }
        }
        return map.get(root.label);
    }
原文地址:https://www.cnblogs.com/aboutblank/p/4379648.html