Why does heap sort have a space complexity of O(1)?

问题

I understand that both quick sort and merge sort need O(n) auxiliary space for the temporary sub-arrays that are constructed, and in-place quick sort requires O(log n) auxiliary space for the recursive stack frames. But for heap sort, it seems like it also has a worst case of O(n) auxiliary space to build the temporary heap, even if the nodes are just pointers to the actual elements.

I came across this explanation :

Only O(1) additional space is required because the heap is built inside the array to be sorted.

But I think this means the original array necessarily already had to be implemented as some sort of tree? If the original array was just a vector, it seems memory for a heap would still have to be allocated.

回答1:

Data in an array can be rearranged into a heap, in place. The algorithm for this is actually surprisingly simple., but I won't go into it here.

For a heap sort, you arrange the data so that it forms a heap in place, with the smallest element at the back (std::make_heap). Then you swap the last item in the array (smallest item in the heap), with the first item in the array (a largish number), and then shuffle that large element down the heap until it's in a new proper position and the heap is again a new min heap, with the smallest remaining element in the last element of the array. (std::pop_heap)

data:         1 4 7 2 5 8 9 3 6 0

make_heap:   [8 7 9 3 4 5 6 2 1 0] <- this is a min-heap, smallest on right

pop_heap(1): [0 7 9 3 4 5 6 2 1 8] <- swap first and last elements
pop_heap(2): 0 [7 9 3 4 8 6 2 5 1] <- shuffle the 8 down the heap

pop_heap(1): 0 1 [9 3 4 8 6 2 5 7] <- swap first and last elements
pop_heap(2): 0 1 [9 7 4 8 6 3 5 2] <- shuffle the 7 down the heap

etc

So no data actually needs to be stored anywhere else, except maybe during the swap step.

For visualization, here's that original heap shown in a standard form

make_heap 
           0
     2           1
  3     4     5     6
               8   7 9
pop_heap
           8                           1                           1
     2           1               2           8               2           5 
  3     4     5     6    ->   3     4     5     6    ->   3     4     8     6 
                   7 9                         7 9                         7 9

回答2:

The cool trick here is since heap is a full binary tree, you can just use a plain array, and for item i, its parent would be the item i/2.

回答3:

HEAP-SORT(A)
{
BUILD-MAX-HEAP(A)
if(i= A.length down to 2)
    exchange A[i] with A[1]
    A.heapSize = A.heapSize-1
    MAX-HEAPIFY(A,1)

}

i/p is stored in array which is passed to heap sort algorithm- HEAP-SORT(A). Array A is interpreted as tree and after BUILD-MAX-HEAP out of it and swapping last element with root and reducing size of heap each time by one and then call MAX-HEAPIFY(A,1) on it.

this all operations we are performing inside that array(A) only - which is given as i/p to algorithm. we are not using any extra space while performing this operation.. So space complexity - O(1).

回答4:

Heap sort is an in-place algorithm; it doesn't require any extra space. Elements are rearranged during each recursive only inside the same array.

It provides a perception that a binary heap or tree is being formed, but in real scenario no tree or heap is being formed.

来源：https://stackoverflow.com/questions/22233532/why-does-heap-sort-have-a-space-complexity-of-o1