这个方程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)
假设基本情况是某个有限常数，它应该在O(2^n)中运行，因为对于每个调用，它都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，通过缓存的值（假设没有副作用），这可以减少到O(n)
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