leetcode @python 133. Clone Graph

题目链接

https://leetcode.com/problems/clone-graph/

题目原文

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.
   Visually, the graph looks like the following:
   

题目大意

对给出的无向图进行复制

解题思路

使用bfs对图进行遍历，对每个顶点进行复制

代码

# Definition for a undirected graph node
# class UndirectedGraphNode(object):
#     def __init__(self, x):
#         self.label = x
#         self.neighbors = []

class Solution(object):
    def cloneGraph(self, node):
        """
        :type node: UndirectedGraphNode
        :rtype: UndirectedGraphNode
        """
        if node == None:
            return None
        queue = []
        map = {}
        nnode = UndirectedGraphNode(node.label)
        queue.append(node)
        map[node] = nnode
        while queue:
            curr = queue.pop()
            for neighbor in curr.neighbors:
                if neighbor not in map:
                    copy = UndirectedGraphNode(neighbor.label)
                    map[curr].neighbors.append(copy)
                    map[neighbor] = copy
                    queue.append(neighbor)
                else:
                    map[curr].neighbors.append(map[neighbor])
        return nnode