Clone Graph
Created: August 11, 2019 by [lek-tin]
Last updated: August 11, 2019
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.
Solution
Depth first search
/*
// Definition for a Node.
class Node {
public int val;
public List<Node> neighbors;
public Node() {}
public Node(int _val,List<Node> _neighbors) {
val = _val;
neighbors = _neighbors;
}
};
*/
class Solution {
private HashMap<Node, Node> visited = new HashMap<Node, Node>();
public Node cloneGraph(Node node) {
return search(node);
}
public Node search(Node node) {
if (node == null) return null;
if (visited.containsKey(node)) return visited.get(node);
Node dup = new Node(node.val, new LinkedList<Node>());
visited.put(node, dup);
for (Node neighbor: node.neighbors) {
Node copy = search(neighbor);
dup.neighbors.add(copy);
}
return dup;
}
}
Breadth first search
/*
// Definition for a Node.
class Node {
public int val;
public List<Node> neighbors;
public Node() {}
public Node(int _val,List<Node> _neighbors) {
val = _val;
neighbors = _neighbors;
}
};
*/
class Solution {
public Node cloneGraph(Node node) {
if (node == null) return null;
Queue<Node> visited = new LinkedList<Node>();
HashMap<Node, Node> map = new HashMap<Node, Node>();
visited.add(node);
Node dup = new Node(node.val, new LinkedList<Node>());
map.put(node, dup);
while (!visited.isEmpty()) {
Node candidate = visited.poll();
for (Node neighbor: candidate.neighbors) {
Node copy = map.get(neighbor);
if (copy == null) {
// If copy doesn't exist, we create a new node with the value of the current neighbor and assign it to copy.
visited.add(neighbor);
copy = new Node(neighbor.val, new LinkedList<Node>());
map.put(neighbor, copy);
}
// map.get(candidate): the duplicate
map.get(candidate).neighbors.add(copy);
}
}
return map.get(node);
}
}
BFS
from collections import deque
"""
# Definition for a Node.
class Node:
def __init__(self, val, neighbors):
self.val = val
self.neighbors = neighbors
"""
class Solution:
def cloneGraph(self, node: 'Node') -> 'Node':
if node == None:
return Node
copy = Node(node.val, [])
clones = {node: copy}
q = deque([node])
while q:
candidate = q.popleft()
clone = clones[candidate]
# BFS
for neighbor in candidate.neighbors:
# If a clone node exists, copy edge by appending the mapped node to clone.neighbors
if neighbor in clones:
clone.neighbors.append(clones[neighbor])
# Otherwise, construct new a node with neighbor.val and an empty list.
else:
newNode = Node(neighbor.val, [])
clones[neighbor] = newNode
clone.neighbors.append(newNode)
q.append(neighbor)
return copy
Mapping shown: