133.Clone Graph (M)

https://leetcode.com/problems/clone-graph/

Given a reference of a node in a connected undirected graph.

Return a deep copy (clone) of the graph.

Each node in the graph contains a value (int) and a list (List[Node]) of its neighbors.

class Node {
    public int val;
    public List<Node> neighbors;
}

Test case format:

For simplicity, each node's value is the same as the node's index (1-indexed). For example, the first node with val == 1, the second node with val == 2, and so on. The graph is represented in the test case using an adjacency list.

An adjacency list is a collection of unordered lists used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.

The given node will always be the first node with val = 1. You must return the copy of the given node as a reference to the cloned graph.

Example 1:

Input: adjList = [[2,4],[1,3],[2,4],[1,3]]
Output: [[2,4],[1,3],[2,4],[1,3]]
Explanation: There are 4 nodes in the graph.
1st node (val = 1)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
2nd node (val = 2)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).
3rd node (val = 3)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
4th node (val = 4)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).

Example 2:

Input: adjList = [[]]
Output: [[]]
Explanation: Note that the input contains one empty list. The graph consists of only one node with val = 1 and it does not have any neighbors.

Example 3:

Input: adjList = []
Output: []
Explanation: This an empty graph, it does not have any nodes.

Constraints:

• The number of nodes in the graph is in the range [0, 100].

• 1 <= Node.val <= 100

• Node.val is unique for each node.

• There are no repeated edges and no self-loops in the graph.

• The Graph is connected and all nodes can be visited starting from the given node.

Solution

深度拷贝

表怎么存储：1.邻接表

Class Graph{

int node;

ArrayList<Node> neighbor;

}

2.邻接矩阵（矩阵里存点）

Version 1:BFS ：时间复杂度O(n)，空间复杂度O(n)

1. 从原图给定的点找到所有点

2. 复制所有的点

3. 复制所有的边

• HashMap里存放的是原节点和复制节点的对应关系，防止建立重复的新节点。

• queue里存放原图节点的BFS遍历。

• 先复制一个头结点，将他们的map存进HashMap.然后头节点进入队列

• 每一次queue出列一个node,然后检查这个node对的所有neighbor,如果没有被visited,就产生一个新的复制节点并把们的关系放进map

• 然后建立这个复制关系的neighbor链表，将每一个neighbor的复制节点都放进当前复制节点的neighbor list.

 public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        if(node==null){
            return null;
        }
        UndirectedGraphNode copynode=new UndirectedGraphNode(node.label);
        HashMap<UndirectedGraphNode,UndirectedGraphNode> map=new HashMap<>();
        Queue<UndirectedGraphNode> queue=new LinkedList<UndirectedGraphNode>();
        //新旧节点的对应
        map.put(node, copynode);
        queue.offer(node);

        while(!queue.isEmpty()){
            UndirectedGraphNode current=queue.poll();
            //copy node
            for(UndirectedGraphNode neighbor:current.neighbors){
                if(!map.containsKey(neighbor)){
                    queue.offer(neighbor);
                    UndirectedGraphNode copyneighbor=new UndirectedGraphNode(neighbor.label);
                    map.put(neighbor,copyneighbor);
                }
                    //connect
                    //这段应该放在外面，避免当前节点连接的是已经复制过的点。
                    UndirectedGraphNode copy=map.get(current);
                    copy.neighbors.add(map.get(neighbor));                                    
                }                        
            }
            return copynode;
     }

与上面同一解法，在处理visited反过来

 public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        if(node==null){
            return null;
        }
        UndirectedGraphNode copynode=new UndirectedGraphNode(node.label);
        HashMap<UndirectedGraphNode,UndirectedGraphNode> map=new HashMap<>();
        Queue<UndirectedGraphNode> queue=new LinkedList<UndirectedGraphNode>();
        //新旧节点的对应
        map.put(node, copynode);
        queue.offer(node);

        while(!queue.isEmpty()){
            UndirectedGraphNode current=queue.poll();
            for(UndirectedGraphNode neighbor:current.neighbors){

                if(map.containsKey(neighbor)){
                    //如果已经存在，直接把他的neighbor复制过去
                    UndirectedGraphNode copy=map.get(current);
                    copy.neighbors.add(map.get(neighbor));
                }
                else{
                    queue.offer(neighbor);
                    UndirectedGraphNode copyneighbor=new UndirectedGraphNode(neighbor.label);
                    map.put(neighbor,copyneighbor);
                    map.get(current).neighbors.add(copyneighbor);
                }
            }                        
        }
        return copynode;
 }

//*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> neighbors;
    public Node() {
        val = 0;
        neighbors = new ArrayList<Node>();
    }
    public Node(int _val) {
        val = _val;
        neighbors = new ArrayList<Node>();
    }
    public Node(int _val, ArrayList<Node> _neighbors) {
        val = _val;
        neighbors = _neighbors;
    }
}
*/

class Solution {
    
    private Map<Node, Node> map;
    public Node cloneGraph(Node node) {
        
        if (node == null) {
            return node;
        }
        
        map = new HashMap<>();
        Set<Node> originalNodeSet = getNodes(node);
        
        // copy Edges
        for(Node original : originalNodeSet)
        {
            for(Node originalNeighbor : original.neighbors)
            {
                map.get(original).neighbors.add(map.get(originalNeighbor));
            }
        }
        return map.get(node);
        
    }
    
    // Copy Node 
    public Set<Node> getNodes(Node node)
    {
        Set<Node> res = new HashSet<>();
        Queue<Node> queue = new LinkedList<Node>();
        
        map.put(node, new Node(node.val));
        queue.offer(node);
        
        while(!queue.isEmpty())
        {
            Node current = queue.poll();
            res.add(current);
            for(Node adj: current.neighbors)
            {
                if(!res.contains(adj))
                {
                    map.put(adj, new Node(adj.val));
                    queue.offer(adj);
                }
            }
        }
        return res;
    }
    
}

九章answer : http://www.jiuzhang.com/solutions/clone-graph/

http://www.cnblogs.com/springfor/p/3874591.html

Version 2：DFS(Time exceed)

 public UndirectedGraphNode cloneGraph2(UndirectedGraphNode node){
     if(node==null){
         return null;
     }
     HashMap<UndirectedGraphNode, UndirectedGraphNode> hm=new HashMap<>();
     UndirectedGraphNode head=new UndirectedGraphNode(node.label);
     hm.put(node, head);
     dfs(hm,node);
     return head;         
 }

 public void dfs(HashMap<UndirectedGraphNode, UndirectedGraphNode> hm, UndirectedGraphNode node){
     if(node==null){
         return;
     }
     for(UndirectedGraphNode aneighbor: node.neighbors){
         if(!hm.containsKey(aneighbor)){
             UndirectedGraphNode copynode=new UndirectedGraphNode(aneighbor.label);
             hm.put(aneighbor, copynode);
             hm.get(node).neighbors.add(copynode);
         }else{
             dfs(hm,aneighbor);
         }
     }
 }