Looking at the original question at the beginning at this thread, ... I
really can't rephrase it to the one you just mentioned. All it needs is the
"matrix", ... who cares about printing it out.
On 12/5/06, Vikram Venkatesan <[EMAIL PROTECTED]> wrote:
>
> Hi,
> I think the question is not to FORM A MATRIX, but to PRINT the
> elements in the form of spiral matrix, where the problem of SEQUENTIAL
> OUTPUT comes...
>
>
> -Vikram
> On 12/6/06, Gene <[EMAIL PROTECTED]> wrote:
> >
> >
> >
> > hijkl wrote:
> > > this question was asked by Google..
> > > "Write a program of spiral matrix"
> > > ie. it takes inputs and puts in to matrix as a spiral..example.
> > > given : 3 X 4 matrix
> > > your input in this order : 1 5 8 9 10 7 4 8 0 2 3 6
> > > will generate following matrix
> > >
> > > 1 5 8 9
> > > 2 3 6 10
> > > 0 8 4 7
> > >
> > > and big O notation for this..
> >
> > O(k) where k = mn is the number of inputs and the matrix is m x n.
> >
> > Let i0 and i1 be top and bottom rows of the unfilled portion; j0 and j1
> > are left- and rightmost unfilled columns. Then just walk around the
> > edge of the unfilled space, updating as you go. You're done if there
> > are no unfilled rows or columns left.
> >
> > i0 = j0 = 0; i1 = m - 1; j1 = n - 1;
> > for (;;) {
> > if (i0 > i1) break;
> > for (i = i0++, j = j0; j <= j1; j++) a[i][j] = next_input();
> > if (j0 > j1) break;
> > for (i = i0, j = j1--; i <= i1; i++) a[i][j] = next_input();
> > if (i0 > i1) break;
> > for (i = i1--, j = j1; j >= j0; j--) a[i][j] = next_input();
> > if (j0 > j1) break;
> > for (i = i1, j = j0++; i >= i0; i--) a[i][j] = next_input();
> > }
> >
> >
> >
> >
> >
>
> >
>
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