動機

email會在不同的人之間重複，這樣要讓這兩個人合起來

這裡的任務是

  1. 如何找出有重複的人
  2. 如何合起來

Problem

Given a list of accounts where each element accounts[i] is a list of strings, where the first element accounts[i][0] is a name, and the rest of the elements are emails representing emails of the account.

Now, we would like to merge these accounts. Two accounts definitely belong to the same person if there is some common email to both accounts. Note that even if two accounts have the same name, they may belong to different people as people could have the same name. A person can have any number of accounts initially, but all of their accounts definitely have the same name.

After merging the accounts, return the accounts in the following format: the first element of each account is the name, and the rest of the elements are emails in sorted order. The accounts themselves can be returned in any order.

 

Example 1:

Input: accounts = [[John,johnsmith@mail.com,john_newyork@mail.com],[John,johnsmith@mail.com,john00@mail.com],[Mary,mary@mail.com],[John,johnnybravo@mail.com]]Output: [[John,john00@mail.com,john_newyork@mail.com,johnsmith@mail.com],[Mary,mary@mail.com],[John,johnnybravo@mail.com]]Explanation:The first and third John's are the same person as they have the common email johnsmith@mail.com.The second John and Mary are different people as none of their email addresses are used by other accounts.We could return these lists in any order, for example the answer [['Mary', 'mary@mail.com'], ['John', 'johnnybravo@mail.com'], ['John', 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com']] would still be accepted.

Example 2:

Input: accounts = [[Gabe,Gabe0@m.co,Gabe3@m.co,Gabe1@m.co],[Kevin,Kevin3@m.co,Kevin5@m.co,Kevin0@m.co],[Ethan,Ethan5@m.co,Ethan4@m.co,Ethan0@m.co],[Hanzo,Hanzo3@m.co,Hanzo1@m.co,Hanzo0@m.co],[Fern,Fern5@m.co,Fern1@m.co,Fern0@m.co]]Output: [[Ethan,Ethan0@m.co,Ethan4@m.co,Ethan5@m.co],[Gabe,Gabe0@m.co,Gabe1@m.co,Gabe3@m.co],[Hanzo,Hanzo0@m.co,Hanzo1@m.co,Hanzo3@m.co],[Kevin,Kevin0@m.co,Kevin3@m.co,Kevin5@m.co],[Fern,Fern0@m.co,Fern1@m.co,Fern5@m.co]]

 

Constraints:

  • 1 <= accounts.length <= 1000
  • 2 <= accounts[i].length <= 10
  • 1 <= accounts[i][j] <= 30
  • accounts[i][0] consists of English letters.
  • accounts[i][j] (for j > 0) is a valid email.

Sol

這題當初卡在是要用email還是人去認共同祖先

但到後面(寫文章的現在)才意識到，要合起來就是共同祖先阿!! 所以直接看人就好(用人去在union find中合併)

而email就是紀錄有沒有被重複登記即可

def parent(arr,i):
    while arr[i] != i:
        i = arr[i]
    return i
def union(arr,i,j):
    a = parent(arr,i)
    b = parent(arr,j)
    arr[a] = b
class Solution:
    def accountsMerge(self, ac: List[List[str]]) -> List[List[str]]:
        emails = {}
        arr = list(range(0,len(ac)))
        i = 0
        for l in ac:
            for e in l[1:]:
                if e not in emails:
                    emails[e] = i
                else:
                    union(arr,i,emails[e])
            i += 1
        tbl = {}
        ks = sorted(emails.keys())
        for e in ks:
            i = parent(arr,emails[e])
            if i not in tbl:
                tbl[i] = [e]
            else:
                tbl[i].append(e)
        ret = []
        for k in tbl:
            ret.append([ac[k][0]] + tbl[k])
        return ret