Both problems are essentially the same. You need to traverse the matrix in a
spiral order and has a pretty straight forward solution.
-Vijju
On 12/6/06, Lego Haryanto <[EMAIL PROTECTED]> wrote:
>
> 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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