133. Clone Graph (Medium)
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 val (int
) and a list (List[Node]
) of its neighbors.
Example:
Input: {"$id":"1","neighbors":[{"$id":"2","neighbors":[{"$ref":"1"},{"$id":"3","neighbors":[{"$ref":"2"},{"$id":"4","neighbors":[{"$ref":"3"},{"$ref":"1"}],"val":4}],"val":3}],"val":2},{"$ref":"4"}],"val":1} Explanation: Node 1's value is 1, and it has two neighbors: Node 2 and 4. Node 2's value is 2, and it has two neighbors: Node 1 and 3. Node 3's value is 3, and it has two neighbors: Node 2 and 4. Node 4's value is 4, and it has two neighbors: Node 1 and 3.
Note:
- The number of nodes will be between 1 and 100.
- The undirected graph is a simple graph, which means no repeated edges and no self-loops in the graph.
- Since the graph is undirected, if node p has node q as neighbor, then node q must have node p as neighbor too.
- You must return the copy of the given node as a reference to the cloned graph.
Solutions
class Solution {
// record the visited node to avoid multi times access
Map<Node, Node> visited = new HashMap<>();
public Node cloneGraph(Node node) {
if (node == null) {
return null;
}
if (visited.containsKey(node)) {
return visited.get(node);
}
Node newNode = new Node(node.val, new ArrayList<>());
visited.put(node, newNode);
if (node.neighbors == null) {
return newNode;
}
for (Node n : node.neighbors) {
// DFS technique to go through and clone all children nodes
Node xn = cloneGraph(n);
newNode.neighbors.add(xn);
}
return newNode;
}
}