[一起来刷leetcode吧][4]--No.133 Clone Graph

最新推荐文章于 2022-03-11 16:00:26 发布

原创最新推荐文章于 2022-03-11 16:00:26 发布 · 151 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#[python #leetcode #os]

leetcode 同时被 2 个专栏收录

35 篇文章

订阅专栏

一起来刷leetcode吧

34 篇文章

订阅专栏

本文解析了LeetCode上的第133题Clone Graph，详细介绍了如何使用深度优先搜索（DFS）来克隆一个无向图，并通过递归实现节点的复制及邻居的连接。

这篇文章是程序自动发表的,详情可以见这里
href="http://ounix1xcw.bkt.clouddn.com/github.markdown.css" rel="stylesheet">

这是leetcode的第133题--Clone Graph

题目 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 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:

思路 dfs,以及记录哪些被访问了，而且要注意复制的时候用self.nd里的结点

show me the 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 __init__(self):
        self.nd={}
    def cloneGraph(self, node):
        if not node :return None
        cur = UndirectedGraphNode(node.label)
        self.nd[cur.label]=cur
        for i in node.neighbors:
            if i.label==node.label:
                cur.neighbors.append(cur)
                continue
            if i.label in self.nd:
                cur.neighbors.append(self.nd[i.label])
            else:
                ret = self.cloneGraph(i)
                if ret!=None :cur.neighbors.append(ret)
        return cur