[leecode]Clone Graph

Clone Graph

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
         / 
         \_/

算法分析:

BFS的变种,维护一个map来记录已经生成的copy node,如果已经生成,则直接从map中取就行了,否则需要new出一个新的。

维护一个set来记录在queue中,或者已经出queue的点(其实没有必要用set的,直接可以在queue里查询的,但是考虑到效率,还是用了set,而且只记录label,空间也很省)

代码如下:

 1 public class Solution {
 2     public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
 3         if(node == null) return null;
 4         Queue<UndirectedGraphNode> q = new LinkedList<UndirectedGraphNode>();
 5         Set<Integer> set = new HashSet<Integer>();
 6         Map<Integer,UndirectedGraphNode> map = new HashMap<Integer,UndirectedGraphNode>();
 7         set.add(node.label);
 8         q.offer(node);
 9         map.put(node.label,new UndirectedGraphNode(node.label));
10         while(!q.isEmpty()){
11             UndirectedGraphNode tem = q.poll();
12             UndirectedGraphNode copy = map.get(tem.label);
13             List<UndirectedGraphNode> neighbors = tem.neighbors;
14             for(UndirectedGraphNode neighbor : neighbors){
15                 if(!set.contains(neighbor.label)){
16                     set.add(neighbor.label);
17                     q.offer(neighbor);
18                 }
19                 if(!map.containsKey(neighbor.label)){
20                     UndirectedGraphNode copyNeighbor = new UndirectedGraphNode(neighbor.label);
21                     map.put(neighbor.label,copyNeighbor);
22                     copy.neighbors.add(copyNeighbor);
23                 }else{
24                     copy.neighbors.add(map.get(neighbor.label));
25                 }
26             }
27         }
28         return map.get(node.label);
29     }
30 }
原文地址:https://www.cnblogs.com/huntfor/p/3935835.html