[英]What is the running time complexity of this for loop
for (int p = t; p > 0; p >>= 1) {
for (int i = 0; i < n - p; ++i) {
if ((i & p) != 0) {
sort2(a, i, i+p);
}
}
for(int q = t; q > p; q >>= 1) {
for(int i = 0; i < n- q; ++i) {
if ((i & p) != 0) {
sort2(a, i+p, i+q);
}
}
}
}
Here n
is some positive integer and t
is greater than n/2
, but not equal to n
. 这里
n
是一些正整数,并且t
大于n/2
,但不等于n
。
As per my understanding, the inner for loop runs for (np)
times but I could not figure out the outer for loop. 根据我的理解,内部for循环运行
(np)
次,但是我无法弄清楚外部for循环。
I tried finding it as below: 我尝试找到它,如下所示:
If t=64
and n=100
, it takes the binary value of p
which is equal to 64
and so p=1000000
base 2 . 如果
t=64
且n=100
,则采用p
的二进制值,该值等于64
,因此p=1000000
2 p=1000000
底 。
I understand that every time it reduces by one digit and it executes a total of 7 times in this case. 我知道,每次它减少一位数,在这种情况下,它总共执行7次。 I somehow couldn't figure out general time.
我不知道怎么算一般时间。
Also, my understanding is that the 3rd for loop ie 另外,我的理解是第三个for循环
for(int q = t; q > p; q >>= 1)
doesn't execute at all because the condition q>p
doesn't satisfy as p=q=t
. 因为条件
q>p
不满足,因为p=q=t
所以根本不执行。
Is this correct? 这个对吗? I am just starting out with algorithms.
我只是从算法开始。
For this: Complexity would be BigO( log(t)(nt)log(t) )
为此:复杂度为
BigO( log(t)(nt)log(t) )
Excluding the sort2
function complexity inside the for
loop. 不包括
for
循环内的sort2
函数复杂性。
Explaination: 说明:
log(p)+1
[Each time it is right shifting the bit by 1 and going for greater than 0
], so for t = 64, the loop will go as [64, 32, 16, 8, 4, 2, 1]. log(p)+1
[每次将位右移1并greater than 0
],因此对于t = 64,循环将变为[64,32,16,8, 4,2,1]。 greater of ( O(np), O((nt)log(t)) )
greater of ( O(np), O((nt)log(t)) )
Execution 执行
for the second inner loop having nested for loop with the most outer loop: 对于第二个内部循环,嵌套了for循环和最外部的循环:
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