Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

For example,
Given nums = [0, 1, 3] return 2.

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

思路

找到连续数组当中缺失的那个数。

要想实现O(1)的空间复杂度，可以利用等差数列公式，将数组的和与等差数列的和进行比较即可。

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class Solution(object):
    def missingNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        add = 0
        length = len(nums)
        for i in nums:
            add += i
        
        should = ( (1 + length) * length ) / 2
        
        return should - add