961. N-Repeated Element in Size 2N Array

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class Solution:     def repeatedNTimes(self, A: List[int]) -> int:         if len(A)&1 > 0:             return -1         aa = set(A)         bb = {}         for a in aa:             bb[a] = 0         for i in A:             bb[i] += 1             if bb[i]> 1 :                 result = i                 break         return result

　　60ms,14.1M

 

优化一：

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class Solution:     def repeatedNTimes(self, A: List[int]) -> int:         return (sum(A)-sum(set(A)))//(len(A)//2-1)

　　56ms，13.8M

 

优化二：

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class Solution:     def repeatedNTimes(self, A: List[int]) -> int:         for i in A:             if A.count(i)>1:                 return i         return -1

　　52ms，14.3M