Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

Example 1:

Input:
[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]

Example 2:

Input:
[
  [1, 2, 3, 4],
  [5, 6, 7, 8],
  [9,10,11,12]
]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]

 

 

Solution:

We can simulate the spiral route using a while loop.

At the first edge, we add a complete row into result list;

In two middle edges, we add row/ column which starts from the second index into result list;

At the last edge, we add the column which starts from the second index and end on (last – 1) index into the result list.

One time after the while loop, we shrink each edge of the size.

 

class Solution {
    public List < Integer > spiralOrder(int[][] matrix) {
        List<Integer> res = new ArrayList<>();
        
        if (matrix.length == 0) {
            return res;
        }
        
        int rStart = 0;
        int rEnd = matrix.length - 1;
        int cStart = 0;
        int cEnd = matrix[0].length - 1;
        
        while (rStart <= rEnd && cStart <= cEnd) {
            for (int c = cStart; c <= cEnd; c++) {
                res.add(matrix[rStart][c]);
            }
            for (int r = rStart + 1; r <= rEnd; r++) {
                res.add(matrix[r][cEnd]);
            }
            
            if (rStart < rEnd && cStart < cEnd) {
                for (int c = cEnd - 1; c >= cStart; c--) {
                    res.add(matrix[rEnd][c]);
                }
                for (int r = rEnd - 1; r > rStart; r--) {
                    res.add(matrix[r][cStart]);
                }
            }
            
            rStart++;
            cStart++;
            rEnd--;
            cEnd--;
        }
        
        return res;
    }
}

Time Complexity: $N$

Space Complexity: $,\; the\; information\; stored\; in\; result\; list.$