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**Unformatted text preview: **htly larger dataset.
3, 7, 4, 2, 7, 8, 5, 1 (a duplicate too.)
3 8 values, so the last parent is nlast = 3.
nlast
The subtree with that as root is a heap.
Proceed to node 2.
A non-heap. Parent 4 has children 8 and 5.
nonSwap 4 and 8.
3 0 0 7 4 7 1 2 1 4
2 2 7 8 5 2 7 8 5 3 4 5 6 3 4 5 6 1
7 1
node number 7 18 Of course, such a move may cause a subtree
to be no longer a heap.
That's why we have to recheck all subtrees
recursively. (Not needed here.) So we move to node 1. It's a heap.
And finally to the root.
We swap the 3 for the larger of its children - the 8. 3 3 0 0 7 8 7 1 2 1 2 7 4 5 3 4 5 6 8
2 2 5 4 5 6 1 7 4 3 1 7 7 Now node 2 is not a heap.
Swap 3 and 5, the larger of its children. Now node 2 is not a heap.
Swap 3 and 5, the larger of its children.
And we’re finished.
We have a heap. 8 8 0 0 7 3 7 1 2 1 5
2 2 7 4 5 2 7 4 3 3 4 5 6 3 4 5 6 1 1 7 7 Let's see what the code looks like.
Assume the following data items (for an integer
data set).
int n, nlast;
nlast;
int* value;
// dynamic array Here's the code to make a heap provided that
each sub...

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