3。leetcode在2N的陣列中找出N的沖服元素

摘要： 1.題目： 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 eleme...

1.題目：

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

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. 優秀解法：

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

有重複的和減去沒有重複的和 再除以長度除以2再減1就是重複的項。