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.
function spiralOrder(matrix) {
const result = [];
let top = 0;
let bottom = matrix.length - 1;
let left = 0;
let right = matrix[0].length - 1;
while (top <= bottom && left <= right) {
for (let i = left; i <= right; i++) {
result.push(matrix[top][i]);
}
top++;
for (let i = top; i <= bottom; i++) {
result.push(matrix[i][right]);
}
right--;
if (top <= bottom) {
for (let i = right; i >= left; i--) {
result.push(matrix[bottom][i]);
}
bottom--;
}
if (left <= right) {
for (let i = bottom; i >= top; i--) {
result.push(matrix[i][left]);
}
left++;
}
}
return result;
}
Follow-up:
What if the matrix is not a square matrix?
