描述
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?
解析
根据题意肯定用公式计算方法最快，就是把 M 中，每个位置上的数字变成它周围数字（包括自身）总和的平均数，如果是角落的或者边缘的数字，那就只计算若干个数字的平均值，时间复杂度为 O(N) ，公式计算为 O(1)，但是求和过程为 O(N)，空间复杂度为 O(1)。
解答
class Solution(object):
def missingNumber(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
N = len(nums)
total = sum(nums)
tmp = N*(N+1)/2
return tmp-total
运行结果
Runtime: 112 ms, faster than 87.25% of Python online submissions for Missing Number.
Memory Usage: 12.8 MB, less than 30.79% of Python online submissions for Missing Number.
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