[leetcode: Python]268. Missing Number

题目：
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?
题意：
给定一个数组包含n个数，这n个数取自0到n的序列，求失踪的那个数。
要求：
时间复杂度O(n)，空间复杂度O(1)

方法一：性能88ms
sum1 为0到n的和，sum2为给定数组的和，求两者的差异。考虑到数组排序后，nums[0] != 0的情况。

class Solution(object):    def missingNumber(self, nums):        """        :type nums: List[int]        :rtype: int        """        nums.sort()        if nums == []:            return 0        if nums[0] != 0:            return 0        s = sum(range(nums[-1]+1))        for i in nums:            s -= i        if s == 0:            return nums[-1] + 1        return s

方法二：性能45ms

import mathclass Solution(object):    def missingNumber(self, nums):        """        :type nums: List[int]        :rtype: int        """        num_total = sum(nums)        n = len(nums)        total = (n * (n + 1)) / 2        return total - num_total