Clone an undirected graph. Each node in the graph contains a label
and a list of its neighbors
.
OJ‘s undirected graph serialization:
Nodes are labeled uniquely.
We use #
as a separator for each node, and ,
as
a separator for node label and each neighbor of the node.
As an example, consider the serialized graph {0,1,2#1,2#2,2}
.
The graph has a total of three nodes, and therefore contains three parts as separated by #
.
- First node is labeled as
0
. Connect node0
to
both nodes1
and2
. - Second node is labeled as
1
. Connect node1
to
node2
. - Third node is labeled as
2
. Connect node2
to
node2
(itself), thus forming a self-cycle.
There are two solutions for this problem both based on BFS
one iterative another recursive.
The time complexity if O(V+E) and space is the same.
we need to maintain the que.
1. Recursive:
# Definition for a undirected graph node # class UndirectedGraphNode: # def __init__(self, x): # self.label = x # self.neighbors = [] class Solution: # @param node, a undirected graph node # @return a undirected graph node def __init__(self): self.dict={None:None} def cloneGraph(self, node): if node==None: return None head=UndirectedGraphNode(node.label) self.dict[node]=head for n in node.neighbors: if n in self.dict: head.neighbors.append(self.dict[n]) else: neigh=self.cloneGraph(n) head.neighbors.append(neigh) return head
For iterative solution.
It looks more intuitive.
# Definition for a undirected graph node # class UndirectedGraphNode: # def __init__(self, x): # self.label = x # self.neighbors = [] class Solution: # @param node, a undirected graph node # @return a undirected graph node def cloneGraph(self, node): if node==None: return node que=[node] head=UndirectedGraphNode(node.label) dict={node:head} while que: curr=que.pop(0) for n in curr.neighbors: if n in dict: dict[curr].neighbors.append(dict[n]) else: que.append(n) copy=UndirectedGraphNode(n.label) dict[n]=copy dict[curr].neighbors.append(copy) return head
时间: 2024-10-10 15:37:47