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  • 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);
        }
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  • 原文地址:https://www.cnblogs.com/aboutblank/p/4379648.html
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