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Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.
Example 1:
Input: [3,0,1]Output: 2
Example 2:
Input: [9,6,4,2,3,5,7,0,1]Output: 8
Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?思路: 既然列表中的数字是从0开始,一直数到列表的长度为止,中间只缺了一个数,那么用和去减列表的和就可以了.(这道题有bug,输入不符合规范也能算出来答案)想起小时候第一节奥数课背的求和公式:首项加末项乘以项数除以二=.=
class Solution(object): def missingNumber(self, nums): """ :type nums: List[int] :rtype: int """ return (1+len(nums))*(len(nums))//2-sum(nums)
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