Spiral Matrix
Given an m x n matrix, return all elements of the matrix in spiral order.
Constraints:
- m == matrix.length
- n == matrix[i].length
- 1 <= m, n <= 10
- -100 <= matrix[i][j] <= 100
Examples:
Input: [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]
Explanation: The spiral order is: 1 -> 2 -> 3 -> 6 -> 9 -> 8 -> 7 -> 4 -> 5
Solutions
Directional Approach
Time: O(m*n)Space: O(1)
The solution uses four pointers (top, bottom, left, right) to represent the current boundaries of the matrix. It iterates through the matrix in a spiral order by moving the pointers accordingly.
public int[] spiralOrder(int[][] matrix) {
int[] result = new int[matrix.length * matrix[0].length];
int top = 0;
int bottom = matrix.length - 1;
int left = 0;
int right = matrix[0].length - 1;
int index = 0;
while (top <= bottom && left <= right) {
for (int i = left;
i <= right;
i++) {
result[index++] = matrix[top][i];
}
top++;
for (int i = top;
i <= bottom;
i++) {
result[index++] = matrix[i][right];
}
right--;
if (top <= bottom) {
for (int i = right;
i >= left;
i--) {
result[index++] = matrix[bottom][i];
}
bottom--;
}
if (left <= right) {
for (int i = bottom;
i >= top;
i--) {
result[index++] = matrix[i][left];
}
left++;
}
}
return result;
}Follow-up:
What if the matrix is not a square matrix?
