Problem
Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.文章來(lái)源:http://www.zghlxwxcb.cn/news/detail-626162.html
Algorithm
Sum all the numbers as x x x and use n ( n + 1 ) 2 ? x \frac{n(n+1)}{2} - x 2n(n+1)??x.文章來(lái)源地址http://www.zghlxwxcb.cn/news/detail-626162.html
Code
class Solution:
def missingNumber(self, nums: List[int]) -> int:
sum_, n_ = 0, len(nums)
for num in nums:
sum_ += num
return n_ * (n_ + 1) // 2 - sum_
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