Given a non-empty array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
Example 1:
Input: [2,2,1]
Output: 1
Example 2:
Input: [4,1,2,1,2]
Output: 4
class Solution:
     def singleNumber(self, nums):
         dic = {}
         ans = 0
         
         for i in range(len(nums)):
             if str(nums[i]) in dic:
                 dic[str(nums[i])] += 1
             else:
                 dic[str(nums[i])] = 1
         
         for num in dic:
             if dic[num] == 1:
                 ans = int(num)
         
         return ans
         """
         :type nums: List[int]
         :rtype: int
         """

class Solution:
     def singleNumber(self, nums):
         dic = {}
         ans = 0
         
         for i in range(len(nums)):
             if str(nums[i]) in dic:
                 dic[str(nums[i])] += 1
             else:
                 dic[str(nums[i])] = 1
         
         for num in dic:
             if dic[num] == 1:
                 ans = int(num)
         
         return ans
         """
         :type nums: List[int]
         :rtype: int
         """
