1.问题：
In a array A of size 2N, there are N+1 unique elements, and exactly one of these elements is repeated N times.

Return the element repeated N times.
例一：

Input: [1,2,3,3]
Output: 3

例二：

Input: [2,1,2,5,3,2]
Output: 2

注重：

4 <= A.length <= 10000
0 <= A[i] < 10000
A.length is even

1. 我的解法：

class Solution:
    def repeatedNTimes(self, A: List[int]) -> int:
        n = len(A)
        for i in range(0, n):
            if A[i] in (A[i+1:]):
                return A[i]

Runtime: 48 ms, faster than 88.03% of Python3 online submissions for N-Repeated Element in Size 2N Array.
Memory Usage: 14.3 MB, less than 5.12% of Python3 online submissions for N-Repeated Element in Size 2N Array.

1. 优异解法：

 def repeatedNTimes(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        return int((sum(A)-sum(set(A))) // (len(A)//2-1))

有反复的和减去没有反复的和 再除以长度除以2再减1就是反复的项。