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 node0
to both nodes1
and2
. - Second node is labeled as
1
. Connect node1
to node2
. - Third node is labeled as
2
. Connect node2
to node2
(itself), thus forming a self-cycle.
Visually, the graph looks like the following:
1 / / 0 --- 2 / \_/
题解:利用DFS递归的方法克隆图,用hashmap保存原图中的节点和新图中的节点的对应关系。而且利用这个hashmap可以在O(1)的时间内判断一个节点是否被克隆过,并且知道克隆它得到的新节点。
实现一个克隆节点的方法,在这个方法中:
- 判断这个节点是否被clone过了,如果是,返回它对应的clone节点;
- 如果没有被clone过,那么新建节点为该节点的clone节点,并且递归的clone它的邻居节点,放入该节点的neighbors列表里面;
代码如下:
1 /** 2 * Definition for undirected graph. 3 * class UndirectedGraphNode { 4 * int label; 5 * List<UndirectedGraphNode> neighbors; 6 * UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); } 7 * }; 8 */ 9 public class Solution { 10 private HashMap<UndirectedGraphNode,UndirectedGraphNode> mapClone = new HashMap<UndirectedGraphNode,UndirectedGraphNode>(); 11 12 private UndirectedGraphNode cloneNode(UndirectedGraphNode node){ 13 if(node == null) 14 return null; 15 //if this node already has a clone 16 if(mapClone.containsKey(node)) 17 return mapClone.get(node); 18 19 //if this node hasn‘t been cloned, we construct a new node copy 20 UndirectedGraphNode copy = new UndirectedGraphNode(node.label); 21 mapClone.put(node, copy); 22 23 //clone all the neighbers of node 24 for(UndirectedGraphNode n:node.neighbors){ 25 copy.neighbors.add(cloneNode(n)); 26 } 27 return copy; 28 } 29 public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) { 30 return cloneNode(node); 31 } 32 }
【leetcode】Clone Graph
时间: 2024-10-10 14:02:34