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此题最巧妙之处在于使用邮箱并查集方式查找对应的账号,通过这种方式把所有连通的账号遍历完成,在DFS的过程中记录答案emails
最坏情况:O(n^2)
class Solution:
def accountsMerge(self, accounts: List[List[str]]) -> List[List[str]]:
from collections import defaultdict
visited_accounts = [False] * len(accounts)
emails_accounts_map = defaultdict(list)
res = []
# Build up the graph.
for i, account in enumerate(accounts):
for j in range(1, len(account)):
email = account[j]
emails_accounts_map[email].append(i)
# DFS code for traversing accounts.
def dfs(i, emails):
if visited_accounts[i]:
return
visited_accounts[i] = True
for j in range(1, len(accounts[i])):
email = accounts[i][j]
emails.add(email)
for neighbor in emails_accounts_map[email]:
dfs(neighbor, emails)
# Perform DFS for accounts and add to results.
for i, account in enumerate(accounts):
if visited_accounts[i]:
continue
name, emails = account[0], set()
dfs(i, emails)
res.append([name] + sorted(emails))
return res