LintCode 137 克隆图(Clone Graph) Python题解
描述
Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors. Nodes are labeled uniquely.
You need to return a deep copied graph, which has the same structure as the original graph, and any changes to the new graph will not have any effect on the original graph.
You need return the node with the same label as the input node.
说明
How we serialize an undirected graph: http://www.lintcode.com/help/graph/
样例
Example1
Input:
{1,2,4#2,1,4#4,1,2}
Output:
{1,2,4#2,1,4#4,1,2}
Explanation:
1------2
\ |
\ |
\ |
\ |
4
同时，类的定义是：
"""
class UndirectedGraphNode:
def __init__(self, x):
self.label = x
self.neighbors = []
"""
Note： 实话说，第一次遇到类似的题目。复制Python的list和dict等时，a=[1,2,3]; b= a; b.append(4); 的时候，a和b都为[1,2,3,4]。这就是浅复制。要想在数据结构上实现深复制，需要把数据结构的各个部分单独复制。本题中，需要把label和neighbor分别复制。
（本题解参考九章）
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"""
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):
root = node #哪个点是入口，待会仍然从这个点开始
if node is None:
return node
# write your code here
nodes = self.getnodes(node)
mapping = {}
for node in nodes:
mapping[node] = UndirectedGraphNode(node.label)
for node in nodes:
#new_node = mapping[node]
for neighbor in node.neighbors:
#new_neighbor = mapping[neighbor]
#new_node.neighbors.append(new_neighbor)
mapping[node].neighbors.append(mapping[neighbor])
return mapping[root]
def getnodes(self,node):
queue = deque([node])
result = set([node])
while(queue):
head = queue.popleft()
for neighbor in head.neighbors:
if(neighbor not in result):
result.add(neighbor)
queue.append(neighbor)
return result