function    [A] = transinplace0(A);
%
%   Input:  A in  csc format
%
%   Output: A is overwritten with A'
%           Warning: The result does not have ordered columns
%
%   computes the transpose of a CSC matrix A
%   (almost) in place 
%
    n = A.n;
    m = A.m;
    nz = A.nz;
    % create one int array J of length nnz

    J = A.i;   %  I - indices assigned to J array does transpose
%
%   reconstruct column indices j so that
%   [A.i, j, A.x]  is in   [i,j,Aij] format
%   On completion  i <-> j  have been swapped
%

    for j = 1:n,
	for (p = A.p(j) : A.p(j+1)-1),
		A.i(p) = j;
	end
    end

    c = zeros (1,n) ;
    for k = 1:nz
	j = J (k) ;
	c (j) = c (j) + 1 ;
    end
    A.p = cumsum ([1 c]) ;
    w = A.p ;
    for col = 1:n
	for k = (A.p (col)) : (A.p (col+1)-1)

            while (col ~= J(k))
		% the kth triplet is out of place
		ik = A.i (k) ;
		jk = J (k) ;
		xk = A.x (k) ;
		% find out where it should go in the final result
		p = w (jk) ;
		w (jk) = w (jk) + 1 ;
		% swap
		A.i (k) = A.i (p) ;
		J (k) = J (p) ;
		A.x (k) = A.x (p) ;
		A.i (p) = ik ;
		J (p) = jk ;
		A.x (p) = xk ;
		% now the kth triplet might still be out of place, but the
		% pth triplet is in its final position.  The pth triplet will
		% not be considered again as a "p"th triplet, since it is
		% inside the sorted range for its column.  Thus, we must
		% eventually place the kth triplet in its final place.
		% The algorithm will progress to the final solution: QED.

	    end

	end
    end