Let k=no. of rows
If k=odd
mirror the matrix about the ((k/2)+1) th row..
else start from k/2 and (k/2 +1 )th row, swap the two rows.. then
k/2-1 and k/2 +2 and swap and soon...
OR
maintain array as a pointer table.. keep on changing the base address
to which a[i points to..
Let *a[]={{1,1,0},{2,3,4},{4,57,8},{1,2,5}};
Take pointer a[0] and swap it with the a[3] address.. lly for a[2]and a[1]...
This should run in O(n) time ,n=no of rows of matrix
I think i made it clear...
This might not be the best solution... i wud love to see better
solutions from anyone..!
Thanks..
On 7/27/11, Deoki Nandan <deok...@gmail.com> wrote:
> rotate a 2D matrix by angle 180
>
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> Deoki Nandan Vishwakarma
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