Leetcode: Clone Graph

Clone Graph

Similar Problems:

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 #.

  1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
  2. Second node is labeled as 1. Connect node 1 to node 2.
  3. Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.
  4. Visually, the graph looks like the following:
  / \
 /   \
0 --- 2
     / \

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## Blog link: https://code.dennyzhang.com/clone-graph
## Basic Ideas: Use a hashmap
##              For each new node, create one.
##              Scan its neighbors, if found new nodes. Create them. And update mapping
##              Update children
## Complexity: Time O(n*n), Space O(n).
# 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 is None: return None
        d = {}
        self.DFSClone(node, d)
        return d[node][0]

    def DFSClone(self, node, d):
        if node is None: return None
        if node not in d:
            newNode = UndirectedGraphNode(node.label)
            d[node] = (newNode, False)
        (newNode, has_finished) = d[node]
        if has_finished is True:
            return newNode

        # mark as DONE to avoid duplicate visits
        d[node] = (newNode, True)
        for neighbor in node.neighbors:
            newNeighbor = self.DFSClone(neighbor, d)
        return newNode

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