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100 lines
4.0 KiB
Python
100 lines
4.0 KiB
Python
#coding: utf-8
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''' mbinary
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#######################################################################
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# File : accountsMerge.py
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# Author: mbinary
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# Mail: zhuheqin1@gmail.com
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# Blog: https://mbinary.xyz
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# Github: https://github.com/mbinary
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# Created Time: 2018-12-18 17:07
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# Description:
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给定一个列表 accounts，每个元素 accounts[i] 是一个字符串列表，其中第一个元素 accounts[i][0] 是 名称 (name)，其余元素是 emails 表示该帐户的邮箱地址。
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现在，我们想合并这些帐户。如果两个帐户都有一些共同的邮件地址，则两个帐户必定属于同一个人。请注意，即使两个帐户具有相同的名称，它们也可能属于不同的人，因为人们可能具有相同的名称。一个人最初可以拥有任意数量的帐户，但其所有帐户都具有相同的名称。
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合并帐户后，按以下格式返回帐户：每个帐户的第一个元素是名称，其余元素是按顺序排列的邮箱地址。accounts 本身可以以任意顺序返回。
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例子 1:
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Input:
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accounts = [["John", "johnsmith@mail.com", "john00@mail.com"], ["John", "johnnybravo@mail.com"], ["John", "johnsmith@mail.com", "john_newyork@mail.com"], ["Mary", "mary@mail.com"]]
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Output: [["John", 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com'], ["John", "johnnybravo@mail.com"], ["Mary", "mary@mail.com"]]
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Explanation:
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第一个和第三个 John 是同一个人，因为他们有共同的电子邮件 "johnsmith@mail.com"。
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第二个 John 和 Mary 是不同的人，因为他们的电子邮件地址没有被其他帐户使用。
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#######################################################################
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'''
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class Solution(object):
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def accountsMerge(self, accounts):
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"""
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:type accounts: List[List[str]]
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:rtype: List[List[str]]
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"""
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mailDic = {}
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for ct,i in enumerate(accounts):
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for j in i[1:]:
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mailDic[j] = (ct,i[0])
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mails = {mail:idx for idx,mail in enumerate(mailDic.keys())}
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mailNum = len(mails)
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self.findSet = [i for i in range(mailNum)]
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for li in accounts:
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n = len(li)
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for i in range(1,n-1):
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for j in range(i+1,n):
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self.union(mails[li[i]],mails[li[j]])
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dic = {}
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mails = {j:i for i,j in mails.items()}
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for i in range(mailNum):
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mail = mails[i]
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n = mailDic[mails[self.find(i)]][0]
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if n in dic:
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dic[n].append(mail)
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else:
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dic[n] = [mail]
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nameId = {i[0]:i[1] for i in mailDic.values()}
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return [[nameId[i]]+sorted(mail) for i,mail in dic.items()]
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def find(self,i):
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if self.findSet[i]!=i:
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n = self.find(self.findSet[i])
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self.findSet[i] = n
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return self.findSet[i]
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def union(self,i,j):
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if i!=j:
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n = self.find(i)
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if n!=self.find(j):
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self.findSet[n] = self.findSet[j]
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class Solution2:
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'''并查映射'''
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def accountsMerge(self, accounts):
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"""
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:type accounts: List[List[str]]
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:rtype: List[List[str]]
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"""
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mailDic = {j:ct for ct,i in enumerate(accounts) for j in i[1:]}
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self.findSet = {i:i for i in mailDic}
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for li in accounts:
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n = len(li)
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for i in range(1,n-1):
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for j in range(i+1,n):
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self.union(li[i],li[j])
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dic = {}
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for mail in self.findSet:
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n = mailDic[self.find(mail)]
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if n in dic:
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dic[n].append(mail)
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else:
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dic[n] = [mail]
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return [[accounts[i][0]]+sorted(mail) for i,mail in dic.items()]
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def find(self,i):
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if self.findSet[i]!=i:
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n = self.find(self.findSet[i])
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self.findSet[i] = n
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return self.findSet[i]
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def union(self,i,j):
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if i!=j:
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n = self.find(i)
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if n!=self.find(j):
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self.findSet[n] = self.findSet[j]
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