10/22 每日一題 (找出array 中重複的)

You are given an integer array nums with the following properties:

• nums.length == 2 * n.
• nums contains n + 1 unique elements.
• Exactly one element of nums is repeated n times.

Return the element that is repeated n times.

 

Example 1:

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

Example 2:

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

Example 3:

Input: nums = [5,1,5,2,5,3,5,4]
Output: 5

 

Constraints:

• 2 <= n <= 5000
• nums.length == 2 * n
• 0 <= nums[i] <= 104
• nums contains n + 1 unique elements and one of them is repeated exactly n times.

class Solution:

    def repeatedNTimes(self, nums: List[int]) -> int:

        repea=[]

        for i in nums:

            if i in repea:

                return i

            else:

                repea.append(i)

---------

class Solution:
    def repeatedNTimes(self, nums: List[int]) -> int:
        data = {}

        for n in nums:
            if n in data: return n
            data[n] = True

---------

        repea=[]
        for i in nums:
            if i in repea:
                return i
            else:
                repea.append(i)