Leetcode[1748] and python3 implementation of unique elements (one traverse, two collections)

# Give you an array of integers nums . The only elements in an array are those that only appear Exactly Once The elements of . 
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# Please return nums The only element of and . 
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# Example 1： 
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# Input ：nums = [1,2,3,2]
# Output ：4
# explain ： The only element is [1,3] , And for 4 .
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# Example 2： 
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# Input ：nums = [1,1,1,1,1]
# Output ：0
# explain ： There is no single element , And for 0 .
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# Example 3 ： 
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# Input ：nums = [1,2,3,4,5]
# Output ：15
# explain ： The only element is [1,2,3,4,5] , And for 15 .
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# Tips ： 
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# 1 <= nums.length <= 100 
# 1 <= nums[i] <= 100 
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# Related Topics Array Hashtable Count 23 0
# leetcode submit region begin(Prohibit modification and deletion)
class Solution:
def sumOfUnique(self, nums: List[int]) -> int:
ss = set()
ss2 = set()
ret = 0
for num in nums:
if num not in ss2:
if num not in ss:
ss.add(num)
ret += num
else:
ret -= num
ss2.add(num)
return ret
# leetcode submit region end(Prohibit modification and deletion)

Add up each element one by one ;

The first set records elements that have already appeared once , Encounter this element again , Need to remove from the accumulation ;

The second set records elements that have appeared twice , Encounter this element again , Just skip .