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.
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©ing 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:
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