1.题目：
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注意：
4 <= A.length <= 10000
0 <= A[i] < 10000
A.length is even- 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.
- 优秀解法：
def repeatedNTimes(self, A):
"""
:type A: List[int]
:rtype: int
"""
return int((sum(A)-sum(set(A))) // (len(A)//2-1))有重复的和减去没有重复的和 再除以长度除以2再减1就是重复的项。
