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

/*
        Iteration Solution:
    */
    public UndirectedGraphNode cloneGraph1(UndirectedGraphNode node) {
        if (node == null) {
            return null;
        }
        
        UndirectedGraphNode root = null;
        
        // store the nodes which are cloned.
        HashMap<UndirectedGraphNode, UndirectedGraphNode> map = 
            new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
        
        Queue<UndirectedGraphNode> q = new LinkedList<UndirectedGraphNode>();
        
        q.offer(node);
        UndirectedGraphNode rootCopy = new UndirectedGraphNode(node.label);
        
        // 别忘记这一行啊。orz..
        map.put(node, rootCopy);
        
        // BFS the graph.
        while (!q.isEmpty()) {
            UndirectedGraphNode cur = q.poll();
            UndirectedGraphNode curCopy = map.get(cur);
            
            // bfs all the childern node.
            for (UndirectedGraphNode child: cur.neighbors) {
                // the node has already been copied. Just connect it and don't need to copy.
                if (map.containsKey(child)) {
                    curCopy.neighbors.add(map.get(child));
                    continue;
                }
                
                // put all the children into the queue.
                q.offer(child);
                
                // create a new child and add it to the parent.
                UndirectedGraphNode childCopy = new UndirectedGraphNode(child.label);
                curCopy.neighbors.add(childCopy);
                
                // Link the new node to the old map.
                map.put(child, childCopy);
            }
        }
        
        return rootCopy;        
    }

2014.12.30 Redo:

/*
        SOLUTION 3: The improved Version.
    */
    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        if (node == null) {
            return null;
        }        
        
        HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
        
        // BUG 1: can't use queue , should use LinkedList.
        Queue<UndirectedGraphNode> q = new LinkedList<UndirectedGraphNode>();
        
        q.offer(node);
        
        // copy the root node. and then put it into the map.
        UndirectedGraphNode nodeCopy = new UndirectedGraphNode(node.label);
        map.put(node, nodeCopy);
        
        while (!q.isEmpty()) {
            UndirectedGraphNode cur = q.poll();
            
            // get out the copy node. We guarantee that it has been copied. Because we always put it into the map before
            // put it into the queue.
            UndirectedGraphNode curCopy = map.get(cur);
            
            // go through all the children node.
            // Line 71: java.util.ConcurrentModificationException. use cur instead of curCopy
            for (UndirectedGraphNode child: cur.neighbors) {
                
                if (map.containsKey(child)) {
                    curCopy.neighbors.add(map.get(child));
                } else {
                    // Only add the child into the map when it is not visited.
                    q.offer(child);
                    
                    // BUG 3: forget to add the new node into the map.
                    UndirectedGraphNode childCopy = new UndirectedGraphNode(child.label);
                    curCopy.neighbors.add(childCopy);
                    map.put(child, childCopy);
                }
            }
        }
        
        return map.get(node);
    }

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。

/*
        Solution 2: Recursion version.
    */
    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        if (node == null) {
            return null;
        }
        
        return rec(node, new HashMap<UndirectedGraphNode, UndirectedGraphNode>());
    }
    
    public UndirectedGraphNode rec(UndirectedGraphNode root, HashMap<UndirectedGraphNode, UndirectedGraphNode> map) {
        // If it has been copied, just return the copy node from the map.
        UndirectedGraphNode rootCopy = map.get(root);
        if (rootCopy != null) {
            return rootCopy;
        }
        
        // if the root is not copied, create a new one.
        rootCopy = new UndirectedGraphNode(root.label);
        map.put(root, rootCopy);
        
        // copy all the child node.
        for (UndirectedGraphNode child: root.neighbors) {
            // call the recursion to create all the children and add the new children to the copy node.
            rootCopy.neighbors.add(rec(child, map));            
        }
        
        return rootCopy;        
    }

2014.12.30 Redo:

public UndirectedGraphNode cloneGraph1(UndirectedGraphNode node) {
        if (node == null) {
            return null;
        }        
        
        return rec(node, new HashMap<UndirectedGraphNode, UndirectedGraphNode>());
    }
    
    // SOLUTION 1:
    // Try to return a copied cloneGraph.
    public UndirectedGraphNode rec(UndirectedGraphNode node, HashMap<UndirectedGraphNode, UndirectedGraphNode> map) {
        // The base case:
        if (map.containsKey(node)) {
            // If the map has been copied, just return the node.
            return map.get(node);
        }
        
        // create a new node.
        UndirectedGraphNode nodeCopy = new UndirectedGraphNode(node.label);
        // BUG 2: should put it into the map first. Because we don't want to copy the same node again in the recursion.
        
        map.put(node, nodeCopy);
        for (int i = 0; i < node.neighbors.size(); i++) {
            // BUG 1: forget a parameter.
            // copy all the children node.
            nodeCopy.neighbors.add(rec(node.neighbors.get(i), map));
        }
        
        
        
        return nodeCopy;
    }

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

Code:

CloneGraph.java