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

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----

/*
// 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 == nullptr)
            return nullptr;
        unordered_map<Node*, Node*> M;
        queue<Node*> Q;
        Q.push(node);
        while (not Q.empty()) {
            auto cur = Q.front();
            Q.pop();
            if (M.find(cur) == M.end()) { // if NOT already cloned
                M[cur] = new Node(cur->val);
                for (auto nei : cur->neighbors) {
                    Q.push(nei);
                }
            }
        }
        // setup the neighbors
        unordered_set<Node*> visited;
        Q.push(node);
        while (not Q.empty()) {
            auto cur = Q.front();
            Q.pop();
            if (visited.find(cur) == visited.end()) {
                visited.insert(cur);
                for (auto nei : cur->neighbors) {
                    M[cur]->neighbors.push_back(M[nei]);
                    Q.push(nei);
                }
            }
        }
        return M[node];
    }
};

----------再仔细考虑，其实可以一个循环走完的。  
  巧妙在Queue中加的都是还没有copy 边的Node*,因此可以直接操作
class Solution {
public:
    Node* cloneGraph(Node* node) {
        if (node == nullptr)
            return nullptr;
        unordered_map<Node*, Node*> M;
        queue<Node*> Q;
        Q.push(node);
        M[node] = new Node(node->val);
        while (not Q.empty()) {
            auto old = Q.front();
            Q.pop();
            for(auto & nei : old->neighbors){
                if(M.find(nei) == M.end()){
                    M[nei] = new Node(nei->val);
                    Q.push(nei); // contains only newly created node
                }
                M[old]->neighbors.push_back(M[nei]);
            }
        }
        return M[node];
    }
};

Mistakes:

 BFS的时候，忘记Q。push（）了。   肏