HashTable 268 Missing Number

solution 1: with O(n) space
    def missingNumber(self, nums):
        return sum([i for i in range(len(nums)+1)]) - sum(nums)
Solution 2: with O（1） space and use zero_points_to_bit to indicate which key has nums[key] == 0
    def missingNumber(self, nums):
        extra_bit,zero_points_to_bit = 0,-1
        for i in range(len(nums)):
            if abs(nums[i]) == len(nums):
                extra_bit = 1
            if abs(nums[i]) < len(nums):
                if nums[abs(nums[i])] == 0:
                    zero_points_to_bit = abs(nums[i])
                nums[abs(nums[i])] = 0 - abs(nums[abs(nums[i])])
        if extra_bit == 0:
            return len(nums)
        for i in range(len(nums)):
            if nums[i] > 0 or nums[i] == 0 and zero_points_to_bit == -1: 
                return i
Solution 3: with O(1) space and replace all keys pointing to zero by using a bigger positive integer to mark
def missingNumber(self, nums):
        extra_bit,count = 0,1
        for x in nums:
            if abs(x) == len(nums):
                extra_bit = 1
            elif abs(x) > len(nums):
                nums[0] = 0 - abs(nums[0])
            else:
                if nums[abs(x)] == 0:
                    nums[abs(x)] = len(nums) + count
                    count += 1
                nums[abs(x)] = 0 - abs(nums[abs(x)])
        if extra_bit == 0:
            return len(nums)
        else:
            for i in range(len(nums)):
                if nums[i] >= 0:
                    return i