While the list length is more than one
Take 2 elements from the head
Select the larger of the two
If the smaller is greater than the largest beaten by the larger
Then set the largest beaten by the larger to the value of
the smaller
Add the larger to the tail of the list
When this completes, you'll have one element containing the largest
and second largest values.
typedef struct
{
unsigned int value;
unsigned int largestBeaten;
} element;
unsigned int secondLargest(queue<element> elements)
{
while(elements.length() > 1)
{
element A = elements.dequeue();
element B = elements.dequeue();
if (A.value < B.value) swap(A,B);
if (A.largestBeaten < B.value) A.largestBeaten = B.value;
elements.enqueue(A);
}
return queue.head().largestBeaten;
}
On Aug 30, 12:53 pm, sangeeta goyal <sangeeta15...@gmail.com> wrote:
> @Don can you give the algorithm for the same??
> how would you implement it??
>
>
>
>
>
>
>
> On Thu, Aug 30, 2012 at 10:03 PM, Don <dondod...@gmail.com> wrote:
> > The second largest element is the largest element beaten by the
> > winner.
> > So if you implement a tournament in which each element keeps track of
> > the largest element it has beaten, you'll get the second largest
> > naturally.
> > Don
>
> > On Aug 29, 9:15 am, Sangeeta <sangeeta15...@gmail.com> wrote:
> > > give the algo or program to find second largest element in a list using
> > > tournament method
>
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