插入STL载体

Question

With a C++ STL vector we are building a vector of N elements and for some reason we chose to insert them at the front of the vector. Every element insertion at the front of a vector forces the shift of all existing elements by 1. This results in (1+2+3+...+N) overall shifts of vector elements, which is (N/2)(N+1) shifts.

My question is how the author came with (1+2+3+...N), I thought it should be 1+1+1..N as we are moving one element at one position to get empty at beginning?

Thanks!

Answer 1

From [vector.modifiers]/2 (which describes vector::insert ):

Complexity: The complexity is linear in the number of elements inserted plus the distance to the end of the vector.

Each time that you add an element the distance to the end of the vector is increased by one.

The first time that you add an element, there is 1 to be inserted and the distance to the end is 0, so the complexity is 1 + 0 = 1. The second time, there is 1 to be inserted, and the distance to the end is 1, so the complexity is 1 + 1 = 2. The third time, the distance to the end is 2, so the complexity is 1 + 2 = 3. This is what creates the 1 + 2 + 3 + ... + N pattern that the author is describing.

Answer 2

At insertion n , there are n elements currently in the vector that needs to be shifted.

vector<int> values;
for (size_t i = 0; i < N; ++i)
{
     //At this point there are `i` elements in the vector that need to be moved
     //to make room for the new element
     values.insert(values.begin(), 0); 
}

Answer 3

The first Value is shifted N-1 times, each time a new value is inserted it has to move. The second value is shifted N-2 times because only N-2 values are added after it. Next value is shifted N-3 and so on. The last value is not shifted.

I don't know why the author speaks about N and not N-1. But the reason for your confusion is, that the author counts the shifts of a single value and you count the amount of shift prozesses involving more than one single value shift.