leetcode 133. Clone Graph 图遍历BFS + 避免循环

喜欢ヅ旅行 2022-06-08 05:57 186阅读 0赞

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
         / \
         \_/

这道题考查的是图的遍历，可以使用BFS或者DFS，不过要避免循环。

代码如下：

import java.util.ArrayList;
    import java.util.HashMap;
    import java.util.LinkedList;
    import java.util.List;
    import java.util.Map;
    import java.util.Queue;
    
    
    /*class UndirectedGraphNode 
    {
          int label;
          List<UndirectedGraphNode> neighbors;
          UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); }
    };
    */
     /*
      * 这个题主要的注意就是使用map映射结点信息
      * */
    public class Solution 
    {
        public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) 
        {
            if(node==null)
                return null;
    
            UndirectedGraphNode newNode = new UndirectedGraphNode(node.label);
            Map<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
            map.put(node, newNode);
    
            Queue<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();
            queue.add(node);
            while(queue.isEmpty()==false)
            {
                UndirectedGraphNode top = queue.poll();
                List<UndirectedGraphNode> list = top.neighbors;
                for(int i=0;i<list.size();i++)
                {
                    UndirectedGraphNode one = list.get(i);
                    if(map.containsKey(one)==false)
                    {
                        UndirectedGraphNode tmp = new UndirectedGraphNode(one.label);
                        map.put(one, tmp);
                        //只有新建的需要加入queue，否者不需要，避免图中的环
                        queue.add(one);
                    }
                    map.get(top).neighbors.add(map.get(one));
                }
            }
            return newNode;
        }
    }