How to solve Spiral Matrix in go - go

Question

https://leetcode.com/problems/spiral-matrix/
golang implement.
the result as follow:
Run Code Status: Runtime Error
Run Code Result: ×
Your input
[]
Your answer
Expected answer
[]
Show Diff
why [] is the test case ,it's just a one-dimensional slice ?
my code is :
func sprial(begin_r, begin_c, row, col int, matrix [][]int) []int {
s := make([]int, col*row, col*row+10)
k := 0
if row == 1 && col == 1 {
s[k] = matrix[begin_r][begin_c]
return s
} else if row == 1 {
return matrix[begin_r][begin_c : col-1]
} else if col == 1 {
return matrix[begin_r : row-1][begin_c]
} else {
for i := begin_c; i < col; i++ {
s[k] = matrix[begin_r][i]
k++
}
for i := begin_r + 1; i < row; i++ {
s[k] = matrix[i][col-1]
k++
}
for i := col - 2; i >= begin_c; i-- {
s[k] = matrix[row-1][i]
k++
}
for i := row - 2; i >= begin_r+1; i-- {
s[k] = matrix[i][begin_c]
k++
}
return s[:k-1]
}
}
func spiralOrder(matrix [][]int) []int {
m := len(matrix)
n := len(matrix[0])
i := 0
j := 0
// var rS []int
k := 0
//s1 := make([]int, m*n, m*n)
var s1 = []int{}
for {
if m <= 0 || n <= 0 {
break
}
s := sprial(i, j, m, n, matrix)
if k == 0 {
s1 = s
} else {
s1 = append(s1, s...)
}
i++
j++
m -= 2
n -= 2
k++
}
return s1
}

Answer 1

func spiralOrder(matrix [][]int) []int {
if len(matrix) == 0 || len(matrix[0]) == 0 {
return nil
}
m, n := len(matrix), len(matrix[0])
next := nextFunc(m, n)
res := make([]int, m*n)
for i := range res {
x, y := next()
res[i] = matrix[x][y]
}
return res
}
func nextFunc(m, n int) func() (int, int) {
top, down := 0, m-1
left, right := 0, n-1
x, y := 0, -1
dx, dy := 0, 1
return func() (int, int) {
x += dx
y += dy
switch {
case y+dy > right:
top++
dx, dy = 1, 0
case x+dx > down:
right--
dx, dy = 0, -1
case y+dy < left:
down--
dx, dy = -1, 0
case x+dx < top:
left++
dx, dy = 0, 1
}
return x, y
}
}
Source: https://github.com/aQuaYi/LeetCode-in-Go/blob/master/Algorithms/0054.spiral-matrix/spiral-matrix.go
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