54. Spiral Matrix

Math

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> result = new ArrayList<>();
        if (matrix == null || matrix.length == 0) return result;

        int top = 0, bottom = matrix.length - 1;
        int left = 0, right = matrix[0].length - 1;

        while (top <= bottom && left <= right) {
            // left -> right
            for (int i = left; i <= right; i++) {
                result.add(matrix[top][i]);
            }
            top++;

            // top -> btn
            for (int i = top; i <= bottom; i++) {
                result.add(matrix[i][right]);
            }
            right--;

            if (top <= bottom) {
                // right -> left
                for (int i = right; i >= left; i--) {
                    result.add(matrix[bottom][i]);
                }
                bottom--;
            }

            if (left <= right) {
                // btn -> top
                for (int i = bottom; i >= top; i--) {
                    result.add(matrix[i][left]);
                }
                left++;
            }
        }

        return result;
    }
}

Remarks:

1. TC: $O(m\times n)$; SC: $O(1)$