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。