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.lengthn == matrix[i].length1 <= m, n <= 10-100 <= matrix[i][j] <= 100func spiralOrder(matrix [][]int) []int {
list := make([]int, 0)
r := 0
c := 0
bigR := len(matrix) - 1
bigC := len(matrix[0]) - 1
for r <= bigR && c <= bigC {
for i := c; i <= bigC; i++ {
list = append(list, matrix[r][i])
}
r++
for i := r; i <= bigR; i++ {
list = append(list, matrix[i][bigC])
}
bigC--
for i := bigC; i >= c && r <= bigR; i-- {
list = append(list, matrix[bigR][i])
}
bigR--
for i := bigR; i >= r && c <= bigC; i-- {
list = append(list, matrix[i][c])
}
c++
}
return list
}