133. Clone Graph ---M

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 val (int) and a list (List[Node]) of its neighbors.

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

 

Test case format:

For simplicity sake, 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.

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.

Example 4:

Input: adjList = [[2],[1]]
Output: [[2],[1]]

 

Constraints:

• 1 <= Node.val <= 100
• Node.val is unique for each node.
• Number of Nodes will not exceed 100.
• There is 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.

A:
 经典DFS题目
这个题目的难点在于 要copy 邻居，导致需要所有的邻居create之后，才能add neighbors， 因此，我们需要先dfs ，再add neighbors

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

class Solution {
public:
    Node* cloneGraph(Node* node) {        
        if(!node)
            return nullptr;
        unordered_map<int, Node*> map; 
        dfs_clone_val(node, map);
        return map[node->val];
    }
private:
    void dfs_clone_val(Node* node,unordered_map<int, Node*> & map){
        if(map.find(node->val) != map.end())
            return;
        Node * newNode = new Node(node->val);
        map[node->val] = newNode;
        for(auto kk : node->neighbors)
            dfs_clone_val(kk, map);
        
        for(auto kk : node->neighbors)
            map[node->val]->neighbors.push_back(map[kk->val]);
    }
};

Mistakes:
 1: when creating a new node, we should add it into the HashMap immediately,   ------------rather than add into hashMap after visiting&copying  his neighbors
 2：总是miss掉很多节点的   问题找到了------------------在create new node的时候，我们也顺便create 了他们的neighbor， 但是， 这些neighbor的 neighbor，并没有被copy过来。
  因此，我们需要check一下。

5:  有个问题，就是说。当我们给定一个node，查询其neighbor的时候，我们只需要建立（或者找到） 新建的，具有相同label的点即可。（这个时候，不需要建立新的点的neighbor）等以后，我们真正visit 其节点的时候，再update neighbor 链表。



-------------------------第3遍----------BFS------------------
唉，这次，3个月没写code，完全忘记了。
唯一想明白的是，由于neighbor的问题， 我们走两遍，第一遍先create all the new nodes, and in the second iteration, we link all the neighbors.
这道题的关键点，在于， 在第二遍遍历的时候，要在每个neighbor List中，加入Set，因为有可能会有2条neighbor指向同一个目标节点。   因此先加入的话，会有重复----

public class Solution {
    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        if(node == null)
            return null;
        // use bfs to traverse the graph, and build the mapping
        Map<Integer,UndirectedGraphNode> map = new HashMap();
        Queue<UndirectedGraphNode> list = new LinkedList();
        list.add(node);
        while( ! list.isEmpty()){
            UndirectedGraphNode oldNode = list.poll();
            UndirectedGraphNode newNode = new UndirectedGraphNode(oldNode.label);
            map.put(oldNode.label,newNode);
            for(UndirectedGraphNode tmp:oldNode.neighbors)
                if( ! map.containsKey(tmp.label))
                    list.add(tmp);
        }
        // add the neighbors
        list.add(node);
        Set<Integer> isVisitSet = new HashSet();
        while( ! list.isEmpty() ){
            UndirectedGraphNode oldNode = list.poll();
            isVisitSet.add(oldNode.label); 
            UndirectedGraphNode newNode = map.get(oldNode.label);
            for(UndirectedGraphNode tmp : oldNode.neighbors){
                newNode.neighbors.add(map.get(tmp.label));
                if(isVisitSet.contains(tmp.label) == false){
                    list.add(tmp);
                    isVisitSet.add(tmp.label);
                }
            }
        }
        return map.get(node.label);
    }
}

Mistakes: