Lintcode 137. Clone Graph (Medium) (Python)

本文介绍了一种用于深拷贝无向图的算法,该算法通过深度优先搜索(DFS)实现。在无向图中,每个节点包含一个标签和邻居节点列表,通过DFS遍历所有节点并创建其副本,确保新旧图完全独立。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Clone Graph

Description:

Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.

How we serialize an undirected graph:

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

1
/ \
/ \
0 — 2
/ \
_/
Example
return a deep copied graph.

Code:

"""
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):
        # write your code here
        if not node:
            return None
        def dfs(input, map):
            if input in map:
                return map[input]
            output = UndirectedGraphNode(input.label)
            map[input] = output
            for i in input.neighbors:
                output.neighbors.append(dfs(i, map))
            return output
        return dfs(node, {})
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值