Java Solution with explanation

• I got it mostly right failing a few cases initially. The tricky part was to realize that you should end the iteration when you have already visited the m*n items of the matrix. The idea is:

As evident with their names, these will set the boundaries of the matrix. So let us do the following in that sequence:

a. Go left--->right and top_row++
b. Go top--->down and right_col--
c. Go right--->left and bottom_row--
d. Go bottom-->up and left_col++

This mostly works but it will fail for matrix such as {{1,2}} as it will work on the same elements twice. So the trick is to check a condition if size(m*n)>list.size(). Code is below:

``````public List<Integer> spiralOrder(int[][] matrix)
{
if(matrix.length==0) {
return Collections.EMPTY_LIST;
}
int top_row = 0;
int left_col = 0;
int bottom_row = matrix.length - 1;
int right_col = matrix[0].length - 1;
// Each complete rotation will change the range to restrict it.
List<Integer> list = new ArrayList<>();
int size = matrix.length * matrix[0].length;
while (top_row <= bottom_row
&& left_col <= right_col)
{
for (int i = left_col; i <= right_col
&& size > list.size(); i++)
{
}
top_row++;
for (int i = top_row; i <= bottom_row
&& size > list.size(); i++)
{
}
right_col--;
for (int i = right_col; i >= left_col
&& size > list.size(); i--)
{
}
bottom_row--;
for (int i = bottom_row; i >= top_row
&& size > list.size(); i--)
{