Question

这个等式的时间复杂度是多少？答案是2 ^ n，但我找不到解决方案。

 t(n)=t(n-1)*t(n-1)

Answer 1

假设基本情况是一些有限常量，它应该在O(2^n)中运行，因为对于每个调用，它会产生两个递归调用。

例如，给定t(1) = constant

的基本情况

t(1) = t(0) * t(0) = constant * constant

t(2) = t(1) * t(1)
  t(1) = t(0) * t(0) = constant * constant
  t(1) = t(0) * t(0) = constant * constant

t(3) = t(2) * t(2)
  t(2) = t(1) * t(1)
    t(1) = t(0) * t(0) = constant * constant
    t(1) = t(0) * t(0) = constant * constant
  t(2) = t(1) * t(1)
    t(1) = t(0) * t(0) = constant * constant
    t(1) = t(0) * t(0) = constant * constant

然而，通过缓存值可以将其减少到O(n)（假设没有副作用）

t(1) = t(0) * t(0) = constant * constant

t(2) = t(1) * t(1)
  t(1) = t(0) * t(0) = constant * constant
  t(1) evaluated in constant time cache lookup

t(3) = t(2) * t(2)
  t(2) = t(1) * t(1)
    t(1) = t(0) * t(0) = constant * constant
    t(1) evaluated in constant time cache lookup
  t(2) evaluated in constant time cache lookup

该等式的时间复杂度t（n）= t（n-1）* t（n-1）

1 个答案: