我的算法的时间和空间复杂性是什么？

Question

当我学习数据结构和算法时，我对计算算法的时间和空间的复杂性感到非常困惑。

这个问题来自 leetcode 。

def product_except_self(nums):
    result = []
    start = 0
    while start != len(nums):
        total = 1
        for i in nums:
            if i != nums[start]:
                total *= i
        result.append(total)
        start += 1
    return result

因此，我的计算复杂性是
时间复杂性：o（n^2）
空间复杂性：o（n）

Answer 1

def product_except_self(nums):
    result = [] // in the worst case you would need to put all array in the result array, what give O(n) space complexity. 
    start = 0
    while start != len(nums): // O(n) complexity
        total = 1
        for i in nums: // One more loop that gave additional O(n) complexity, in result we have O(n^2)
            if i != nums[start]:
                total *= i
        result.append(total)
        start += 1
    return result

是的，你有正确的假设