两段代码的时间复杂度 [英] Time complexity for two pieces of code
问题描述
我们有两段代码:
int a = 3;
while (a <= n) {
a = a * a;
}
还有:
public void foo(int n, int m) {
int i = m;
while (i > 100)
i = i / 3;
for (int k = i ; k >= 0; k--) {
for (int j = 1; j < n; j*=2)
System.out.print(k + "\t" + j);
System.out.println();
}
}
它们的时间复杂度是多少?我认为第一个是:O(logn),因为它以 2 的幂发展到 N.
所以也许是 O(log2n) ?
What is the time complexity of them?
I think that the first one is: O(logn), because it's progressing to N with power of 2.
So maybe it's O(log2n) ?
我相信的第二个是:O(nlog2n),因为它以 2 次跳跃前进,并且还在外循环上运行.
And the second one I believe is: O(nlog2n), because it's progressing with jumps of 2, and also running on the outer loop.
我说得对吗?
推荐答案
我相信,第一个代码将在 O(Log(LogN)) 时间内运行.这样理解很简单
I believe, that first code will run in O(Log(LogN)) time. It's simple to understand in this way
- 在第一次迭代之前,你有 3 次幂 1
- 第一次迭代后,你有 3 次幂 2
- 第二次迭代后,你有 3 的 4 次幂
- 第三次迭代后,你有 3 次幂 8
- 第四次迭代后,你有 3 的 16 次幂等等.
在第二个代码中,第一段代码将在 O(LogM) 时间内工作,因为您每次都将 i 除以 3.第二段代码 C 次(在你的情况下 C 等于 100)将执行 O(LogN) 操作,因为你每次将 j 乘以 2,所以它运行在 O(CLogN) 中,并且你的复杂度为 O(LogM + CLogN))
In the second code first piece of code will work in O(LogM) time, because you divide i by 3 every time. The second piece of code C times (C equals 100 in your case) will perform O(LogN) operations, because you multiply j by 2 every time, so it runs in O(CLogN), and you have complexity O(LogM + CLogN)
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