问题
In a written examination, I meet a question like this:
When a Dynamic Array is full, it will extend to double space, it's just like 2 to 4, 16 to 32 etc. But what's time complexity of putting an element to the array?
I think that extending space should not be considered, so I wrote O(n), but I am not sure.
what's the answer?
回答1:
It depends on the question that was asked.
If the question asked for the time required for one insertion, then the answer is O(n) because big-O implies "worst case." In the worst case, you need to grow the array. Growing an array requires allocating a bigger memory block (as you say often 2 times as big, but other factors bigger than 1 may be used) and then copying the entire contents, which is the n existing elements. In some languages like Java, the extra space must also be initialized.
If the question asked for amortized time, then the answer is O(1). Another way of saying this is that the cost of n adds is O(n).
How can this be? Each addition is O(n), but n of them also require O(n). This is the beauty of amortization. For simplicity, say the array starts with size 1 and grows by a factor of 2 every time it fills, so we're always copying a power of 2 elements. This means the cost of growing is 1 the first time, 2 the second time, etc. In general, the total cost of growing to n elements is TC=1+2+4+...n. Well, it's not hard to see that TC = 2n-1. E.g. if n = 8, then TC=1+2+4+8=15=2*8-1. So TC is proportional to n or O(n).
This analysis works no matter the initial array size or the factor of growth, so long as the factor is greater than 1.
If your teacher is good, he or she asked this question in an ambiguous manner to see if you could discuss both answers.
回答2:
In order to grow the array size you cannot simply "add more to the end" because you will more likely get a "segmentation fault" type of error. So even though as a mean value it takes θ(1) steps because you have enough space, in terms if O notation is O(n) because you have to copy the old array into a new bigger array (for which you allocated memory) and that should take n steps...generally. On the other hand of course that you can copy arrays faster generally because it's just a memory copy from a continuous space and that should be 1 step in the best scenario ,i.e where the page (OS) can take the whole array. In the end ... mathematically , even considering that we are making making n / (4096 * 2^10) (4 KB) steps, it still means a O(n) complexity.
来源：https://stackoverflow.com/questions/33331314/dynamic-arrays-time-complexity-of-putting-an-element