leetcode

class :
    def missingNumber(self, nums: List[int]) -> int:
        n = len(nums)
        if (n > 0):
            a = set(nums)
            b = set(list(range(n+1)))
            return list(b - a)[0]

# 数学方法：O(n)时间复杂度，O(1)空间复杂度
class :
    def missingNumber(self, nums: List[int]) -> int:
        real_sum = sum(nums)
        n = len(nums)
        expected_sum = n * (n + 1) // 2
        return expected_sum - real_sum
# 两种解法速度类似，第二中解法所需存储空间更少一些