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 <= 100000 <= A[i] < 10000A.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。