Suppose you've got a single linked list of size N, and you want to perform an operation on every element, beginning at the end.
I've come up with the following pseudocode:
while N > 0
Current = LinkedList
for 0 to N
Current = Current.tail
end
Operation(Current.head)
N := N-1
end
Now I've got to determine which Big-O this algorithm is.
Supposing that Operation() is O(1), I think it's somethi开发者_开发问答ng like this:N + (N-1) + (N-2) + ... + (N-(N-1)) + 1
But I'm not sure what Big-O that actually is. I think it is definitely smaller than O(N^2), but I don't think you can say its O(N) either ...
Your equation is basically that of the triangular numbers, and sums to N(N+1)/2. I'll leave you to determine the O()
from that!
A quicker way to do this is to construct a new list that is the reverse of the original list, and then perform the operations on that.
Your algorithm is O(n^2) as you suggest in your post. You can do it in O(n), though.
It's important to remember that Big-O notation is an upper bound on the algorithm's time complexity.
1+2+3+...+n = n*(n+1)/2 = 0.5*n^2+O(n)
This is O(n^2), and O(n^2) is tight, i.e. there is no lower runtime order that'd contain your runtime.
A faster algorithm that works from front-to-back could have O(n) instead of O(n^2)
Your runtime analysis is correct, the runtime is 1 + 2 + ... + N which is a sum of the arithmetic progression and therefore = (N²-N) / 2.
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