Every node can also have an index field. Keep a global variable index=0
After insertion of every new node do node->index=++index, and during the
output compare the index values, the one which has a smaller value will come
first.
Modify the BST based on the count value obtained i.e
insert the first node with node->index=1 and node->count=1
For the subsequent numbers
void insert(NODE *p)
{
NODE *q;
q=p;
while(q!=NULL)
{
p=q;
if (number==q->info)
{
q->count++;
if (q->count > q->father->count)
swap q and q->father
if (q->count == q->father->count && q->index < q->father->index)
swap q and q->father
}
else if(number < q->info)
q=q->left;
else
q=q->right;
}
if (number>p->info)
insert a new node at the right, node->index=++index,
node->count=1
else
insert a new node at the left, node->index=++index, node->count=1
}
After all the insertion is done now the tree is a BST based on count value.
Perform the inorder traversal and depending on the count value print
node->info count no. of times. The order of occurrence is also preserved.
Please correct me if I'm wrong
On Sat, Jul 3, 2010 at 5:49 PM, jalaj jaiswal <jalaj.jaiswa...@gmail.com>wrote:
> how would we keep account of first in first out then...i mean first index
> should be printed first in case of a tie
>
> On Fri, Jul 2, 2010 at 11:48 PM, Amir hossein Shahriari <
> amir.hossein.shahri...@gmail.com> wrote:
>
>> i think that the best method would be a balanced binary search tree which
>> counts the number of appearances for each element
>> O(nlogn)
>>
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>
>
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> With Regards,
> Jalaj Jaiswal
> +919026283397
> B.TECH IT
> IIIT ALLAHABAD
>
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