LeetCode: Clone Graph 解题报告

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 #.

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
         / \
         \_/

Solution 1:

使用BFS来解决此问题。用一个Queue来记录遍历的节点，遍历原图，并且把复制过的节点与原节点放在MAP中防止重复访问。

图的遍历有两种方式，BFS和DFS

这里使用BFS来解本题，BFS需要使用queue来保存neighbors

但这里有个问题，在clone一个节点时我们需要clone它的neighbors，而邻居节点有的已经存在，有的未存在，如何进行区分？

这里我们使用Map来进行区分，Map的key值为原来的node，value为新clone的node，当发现一个node未在map中时说明这个node还未被clone，

将它clone后放入queue中处理neighbors。

使用Map的主要意义在于充当BFS中Visited数组，它也可以去环问题，例如A--B有条边，当处理完A的邻居node，然后处理B节点邻居node时发现A已经处理过了

处理就结束，不会出现死循环。

queue中放置的节点都是未处理neighbors的节点。

http://www.cnblogs.com/feiling/p/3351921.html

 1 /*
 2         Iteration Solution:
 3     */
 4     public UndirectedGraphNode cloneGraph1(UndirectedGraphNode node) {
 5         if (node == null) {
 6             return null;
 7         }
 8         
 9         UndirectedGraphNode root = null;
10         
11         // store the nodes which are cloned.
12         HashMap map = 
13             new HashMap();
14         
15         Queue q = new LinkedList();
16         
17         q.offer(node);
18         UndirectedGraphNode rootCopy = new UndirectedGraphNode(node.label);
19         
20         // 别忘记这一行啊。orz..
21         map.put(node, rootCopy);
22         
23         // BFS the graph.
24         while (!q.isEmpty()) {
25             UndirectedGraphNode cur = q.poll();
26             UndirectedGraphNode curCopy = map.get(cur);
27             
28             // bfs all the childern node.
29             for (UndirectedGraphNode child: cur.neighbors) {
30                 // the node has already been copied. Just connect it and don't need to copy.
31                 if (map.containsKey(child)) {
32                     curCopy.neighbors.add(map.get(child));
33                     continue;
34                 }
35                 
36                 // put all the children into the queue.
37                 q.offer(child);
38                 
39                 // create a new child and add it to the parent.
40                 UndirectedGraphNode childCopy = new UndirectedGraphNode(child.label);
41                 curCopy.neighbors.add(childCopy);
42                 
43                 // Link the new node to the old map.
44                 map.put(child, childCopy);
45             }
46         }
47         
48         return rootCopy;        
49     }

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2014.12.30 Redo:

 

 1 /*
 2         SOLUTION 3: The improved Version.
 3     */
 4     public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
 5         if (node == null) {
 6             return null;
 7         }        
 8         
 9         HashMap map = new HashMap();
10         
11         // BUG 1: can't use queue , should use LinkedList.
12         Queue q = new LinkedList();
13         
14         q.offer(node);
15         
16         // copy the root node. and then put it into the map.
17         UndirectedGraphNode nodeCopy = new UndirectedGraphNode(node.label);
18         map.put(node, nodeCopy);
19         
20         while (!q.isEmpty()) {
21             UndirectedGraphNode cur = q.poll();
22             
23             // get out the copy node. We guarantee that it has been copied. Because we always put it into the map before
24             // put it into the queue.
25             UndirectedGraphNode curCopy = map.get(cur);
26             
27             // go through all the children node.
28             // Line 71: java.util.ConcurrentModificationException. use cur instead of curCopy
29             for (UndirectedGraphNode child: cur.neighbors) {
30                 
31                 if (map.containsKey(child)) {
32                     curCopy.neighbors.add(map.get(child));
33                 } else {
34                     // Only add the child into the map when it is not visited.
35                     q.offer(child);
36                     
37                     // BUG 3: forget to add the new node into the map.
38                     UndirectedGraphNode childCopy = new UndirectedGraphNode(child.label);
39                     curCopy.neighbors.add(childCopy);
40                     map.put(child, childCopy);
41                 }
42             }
43         }
44         
45         return map.get(node);
46     }

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Solution 2:

同样的，我们也可以使用递归DFS来解决此题，思路与上图一致，但为了避免重复运算产生死循环。当进入DFS时，如果发现map中已经有了拷贝过的值，直接退出即可。

题目虽然简单，但主页君仍然考虑了递归的特性使程序简洁。比如：我们拷贝只拷贝根节点，而子节点的拷贝由recursion来完成，这样可以使程序更加简洁。

注意：要先加入到map,再调用rec ，否则会造成不断地反复拷贝而死循环。

 1 /*
 2         Solution 2: Recursion version.
 3     */
 4     public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
 5         if (node == null) {
 6             return null;
 7         }
 8         
 9         return rec(node, new HashMap());
10     }
11     
12     public UndirectedGraphNode rec(UndirectedGraphNode root, HashMap map) {
13         // If it has been copied, just return the copy node from the map.
14         UndirectedGraphNode rootCopy = map.get(root);
15         if (rootCopy != null) {
16             return rootCopy;
17         }
18         
19         // if the root is not copied, create a new one.
20         rootCopy = new UndirectedGraphNode(root.label);
21         map.put(root, rootCopy);
22         
23         // copy all the child node.
24         for (UndirectedGraphNode child: root.neighbors) {
25             // call the recursion to create all the children and add the new children to the copy node.
26             rootCopy.neighbors.add(rec(child, map));            
27         }
28         
29         return rootCopy;        
30     }

View Code

2014.12.30 Redo:

 1 public UndirectedGraphNode cloneGraph1(UndirectedGraphNode node) {
 2         if (node == null) {
 3             return null;
 4         }        
 5         
 6         return rec(node, new HashMap());
 7     }
 8     
 9     // SOLUTION 1:
10     // Try to return a copied cloneGraph.
11     public UndirectedGraphNode rec(UndirectedGraphNode node, HashMap map) {
12         // The base case:
13         if (map.containsKey(node)) {
14             // If the map has been copied, just return the node.
15             return map.get(node);
16         }
17         
18         // create a new node.
19         UndirectedGraphNode nodeCopy = new UndirectedGraphNode(node.label);
20         // BUG 2: should put it into the map first. Because we don't want to copy the same node again in the recursion.
21         
22         map.put(node, nodeCopy);
23         for (int i = 0; i < node.neighbors.size(); i++) {
24             // BUG 1: forget a parameter.
25             // copy all the children node.
26             nodeCopy.neighbors.add(rec(node.neighbors.get(i), map));
27         }
28         
29         
30         
31         return nodeCopy;
32     }

View Code

 

Ref: http://m.blog.csdn.net/blog/hellobinfeng/17497883

Code:

CloneGraph.java