Medium
Given an m x n
matrix
, return all elements of the matrix
in spiral order.
Example 1:
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]
Example 2:
Input: matrix = [[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]
Constraints:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 10
-100 <= matrix[i][j] <= 100
function spiralOrder(matrix: number[][]): number[] {
const result: number[] = []
let r = 0,
c = 0
let bigR = matrix.length - 1
let bigC = matrix[0].length - 1
while (r <= bigR && c <= bigC) {
for (let i = c; i <= bigC; i++) {
result.push(matrix[r][i])
}
r++
for (let i = r; i <= bigR; i++) {
result.push(matrix[i][bigC])
}
bigC--
for (let i = bigC; i >= c && r <= bigR; i--) {
result.push(matrix[bigR][i])
}
bigR--
for (let i = bigR; i >= r && c <= bigC; i--) {
result.push(matrix[i][c])
}
c++
}
return result
}
export { spiralOrder }