If you meant "rotate a 2D matrix by angle 180" of order n x n
Then you cannot have a O(n) algo, Each of the n^2 elements must be
accessed so you cannot have anything less than n^2
On Jul 27, 10:59 pm, Puneet Gautam <puneet.nsi...@gmail.com> wrote:
> Can anyone give an O(n) solution pls...??
> I think the above code is an O(n^2) solution..
> if i am not wrong...!!!
>
> On 7/27/11, amit <amit.codenam...@gmail.com> wrote:
>
>
>
> > #include <cstdio>
> > #include <algorithm>
> > using namespace std;
>
> > const int MX = 1000;
> > int n, m;
> > int a[MX][MX];
>
> > int main() {
> > scanf("%d%d", &n, &m);
> > for(int i = 0; i < n; i++)
> > for(int j = 0; j < m; j++)
> > scanf("%d", &a[i][j]);
>
> > for(int i = 0; i < n/2; i++)
> > for(int j = 0; j < m; j++)
> > swap(a[i][j], a[n-i-1][m-j-1]);
> > if(n&1)
> > for(int j = 0; j < m/2; j++)
> > swap(a[n/2][j], a[n/2][m-j-1]);
>
> > for(int i = 0; i < n; i++) {
> > for(int j = 0; j < m; j++)
> > printf("%d ", a[i][j]);
> > printf("\n");
> > }
> > }
>
> > On Jul 27, 7:54 pm, Anika Jain <anika.jai...@gmail.com> wrote:
> >> is it lyk for {1,2,3
> >> 4,5,6,
> >> 7,8,9}
> >> to be {3,2,1,
> >> 6,5,4,
> >> 9,8,7} ??
>
> >> On Wed, Jul 27, 2011 at 9:37 AM, Deoki Nandan <deok...@gmail.com> wrote:
> >> > rotate a 2D matrix by angle 180
>
> >> > --
> >> > **With Regards
> >> > Deoki Nandan Vishwakarma
>
> >> > *
> >> > *
>
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