如何在 Python 3 中定义一个函数来对数组进行排序并返回只存在一次而不是两次的数字

Question

in this code I want to do the following 1- sort the array 2- Iterate over it to find an element that exists only once while every other element exist twice. e,g [2,1,3,2,3], answer is 1 3- In iteration I do the following ->If reached last element , then last element is the wanted ->If an element i is not equal to element i+1, then element i is the wanted 4- at the end return -1 if not found

this is my code, please help me, I am new in Python

class Main:
    def singleNumber(self, nums: List[int]) -> int:
        sorted(nums)
        l =len(nums)
        while i<l:
            if i+1==l:
                re=nums(i)
            if nums(i)!=nums(i+1):
                re=nums(i)
            else:
                i+=1
            re=-1
        return re

Answer 1

you can use collections.Counter with the built-in function sorted :

from collections import Counter
nums = [2,1,3,2,3]


sorted(k for k, v in Counter(nums).items() if v == 1)

output:

[1]