[英]What should be the time complexity of the given function in terms of Big O?
int find_peak (int n, int A []) {
int left, right, lmid, rmid;
left = 0;
right = n - 1;
while (left + 3 <= right) {
lmid = (2 * left + right) / 3;
rmid = (left + 2 * right) / 3;
if (A[lmid] <= A[rmid])
left = lmid;
else
right = rmid;
}
int x = left;
for (int i = left + 1; i <= right; i ++)
if (A[i] > A[x])
x = i;
return A[x];
}
I was trying to solve this function for its BigO notation, but I am very confused about it.我试图解决这个 function 的 BigO 表示法,但我对此很困惑。 Is it O(log n) or something else?是 O(log n) 还是别的什么? I can somewhat solve it in my head but I can't do it on properly.我可以在脑海中解决它,但我不能正确地做到这一点。
Yes, the loop is cut roughly in half at each iteration是的,循环在每次迭代时大致减半
lmid = (2 * left + right) / 3;
rmid = (left + 2 * right) / 3;
if (A[lmid] <= A[rmid])
left = lmid;
else
right = rmid;
To be precise it's log 1.5 (n) , because the actual length right-left
decreases only by 1 / 3 , not halving the iterations.准确地说,它是log 1.5 (n) ,因为实际right-left
长度只减少了1 / 3 ,而不是将迭代减半。 The complexity is still O(log(n))复杂度仍然是O(log(n))
You can try it here https://onlinegdb.com/rkI5gn3Xd你可以在这里试试 https://onlinegdb.com/rkI5gn3Xd
Thanks to chqrlie for prompting me to give a more detailed answer感谢 chqrlie 提示我给出更详细的答案
The complexity of the above code should be O(log3n) ie.. logn base 3 .上述代码的复杂度应该是O(log3n)即.. logn base 3 。 because in the while loop the value of rmid
always comes out as greater than lmid
.So, for a given value of right ie.. n
, the value of left will be reduced by a multiple of 1/3 at each iteration.因为在while 循环中rmid
的值总是大于lmid
。所以,对于给定的 right 值,即 .. n
,每次迭代时 left 的值将减少 1/3 的倍数。
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