问题描述
我有一个简短的程序在这里：
I have a short program here:
Given any n:
i = 0;
while (i < n) {
k = 2;
while (k < n) {
sum += a[j] * b[k]
k = k * k;
}
i++;
}
这样做的渐近运行时间为O（n日志log n）的。为什么会出现这种情况？我得到了整个计划将至少运行n次。但我不知道如何找到loglogN个。内环取决于K * K，所以它显然将是小于n。而且它只是为N日志N，如果它是K / 2各一次。但如何将你找出答案是日志中记录N +
The asymptotic running time of this is O(n log log n). Why is this the case? I get that the entire program will at least run n times. But I'm not sure how to find log log n. The inner loop is depending on k * k, so it's obviously going to be less than n. And it would just be n log n if it was k / 2 each time. But how would you figure out the answer to be log log n?
推荐答案
有关数学证明，内循环可写成：
For mathematical proof, inner loop can be written as:
T(n) = T(sqrt(n)) + 1
w.l.o.g assume 2 ^ 2 ^ (t-1)<= n <= 2 ^ (2 ^ t)=>
we know 2^2^t = 2^2^(t-1) * 2^2^(t-1)
T(2^2^t) = T(2^2^(t-1)) + 1=T(2^2^(t-2)) + 2 =....= T(2^2^0) + t =>
T(2^2^(t-1)) <= T(n) <= T(2^2^t) = T(2^2^0) + log log 2^2^t = O(1) + loglogn
==> O(1) + (loglogn) - 1 <= T(n) <= O(1) + loglog(n) => T(n) = Teta(loglogn).
然后总时间为O（n loglogn）。
and then total time is O(n loglogn).
为什么内环为T（N）= T（开方（N））+1？第一眼看到时，内循环中断，当k> N，意味着在此之前，K是最少的sqrt（N），或在二级之前，它是最多的sqrt（N），因此运行时间为T（开方（N）） + 2＆GE; T（N）与GE; T（开方（N））+ 1。
Why inner loop is T(n)=T(sqrt(n)) +1?first see when inner loop breaks, when k>n, means before that k was at least sqrt(n), or in two level before it was at most sqrt(n), so running time will be T(sqrt(n)) + 2 ≥ T(n) ≥ T(sqrt(n)) + 1.
这篇关于O（N日志log n）的时间复杂度的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！