我学习了一个测试,发现了这个问题,所以我做了以下,是正确的吗?
while循环运行在O(log3n)。
在for循环运行在约O((N-(一些数学))* log2n),因为我有减号是线性的,我说,整个方法运行在O(nlogn),除非我错了,它的东西,像
O((N-(N / log3n))* log2n)< - 不完全是,但相当,不能真正弄明白
。是负线性这里还是不是?如果没有什么是正确的比戈?
公共无效美孚(INT N,INT M)
{
INT I =米;
而(ⅰ→100)
I = I / 3;
对于(INT K = 1; K> = 0; K--)
{
为(诠释J = 1; J&n种;Ĵ* = 2)
System.out.print(K + t的+ J);
的System.out.println();
}
}
解决方案
在 while循环运行O(logm)
。
在while循环已经结束,我
为< = 100,所以下一个for循环将运行在100倍
内环将运行 O(LOGN)
倍,为外循环的每个迭代。所以,总时间为 O(logm + 100 * LOGN)
= O(logm + LOGN)
。
I'm studying for a test and saw this question , so I did the following, is it correct?
the while loop runs at O(log3n).
the for loop runs at about O((n-(some math))*log2n) so because I have the minus symbol which is linear, I say that the whole method runs at O(nlogn), unless I'm wrong and it's something like
O((n-(n/log3n))*log2n) <- not exactly but quite, can't really figure it out.
is the minus linear here or not? if not what is the correct bigO?
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();
}
}
解决方案
The while loop runs in O(logm)
.
After the while loop has finished, i
is <= 100, so the next for loop will run at most 100 times.
The inner loop will run O(logn)
times, for each iteration of the outer loop. So total time is O(logm + 100*logn)
= O(logm + logn)
.
相关推荐
最新文章