if the matrix is n x n and n is not huge > 5k
Use int arr[((n * n) - n ) / 2 + n];
One approach:
arr[0] = n;
c = 1;
for (i=n; i >= 0; i--)
for (j=0; j < i ; j++)
arr[c++] = get_element_at(n-i, j);
And u can access any element in O(1).
Programmers should realize their critical importance and responsibility in a
world gone digital. They are in many ways similar to the priests and monks
of Europe's Dark Ages; they are the only ones with the training and insight
to read and interpret the "scripture" of this age.
On Thu, Jan 13, 2011 at 3:37 PM, juver++ <[email protected]> wrote:
> One-dimensional array.
> 1 2 4
> 2 3 5
> 4 5 6
> => [1, 2, 3, 4, 5, 6]
> Is your matrix summetric on a diagonal?
>
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