snail, running it with a number argument (for the square size) should look like this:$ ./snail 5 1 2 3 4 5 16 17 18 19 6 15 24 25 20 7 14 23 22 21 8 13 12 11 10 9
So, the solution is straightforward: do alternating row and column sweeps, from the outside going inwards, while incrementing the value to be written into a matrix element.
But first, we have to be able to get the input size of the matrix. For that we'll use the
flag and strconv packages. For the print out, we'll use the fmt package. Here's what I came up with.package main
import (
"fmt"
"flag"
"strconv"
)
func main() {
// Determine the matrix size, n.
var n int = 1
flag.Parse()
if len(flag.Args()) > 0 {
n, _ = strconv.Atoi(flag.Arg(0))
}
if n < 1 {
n = 1
}
// No need to be complicated for n=1
if n == 1 {
fmt.Println(n)
return
}
// Create the matrix
values := make([][]int, n)
for i := 0; i < n; i++ {
values[i] = make([]int, n)
}
// Fill the matrix with values
avalue := 0
for ifill := 0; ifill < n; ifill++ {
fstart := ifill
fend := n-ifill
// upper x forward sweep
for ixf := fstart; ixf < fend-1; ixf++ {
avalue++
values[fstart][ixf] = avalue
}
// right y forward sweep
for iyf := fstart; iyf < fend-1; iyf++ {
avalue++
values[iyf][fend-1] = avalue
}
// lower x backward sweep
for ixb := fend-1; ixb > fstart; ixb-- {
avalue++
values[fend-1][ixb] = avalue
}
// left y backward sweep
for iyb := fend-1; iyb > fstart; iyb-- {
avalue++
values[iyb][fstart] = avalue
}
}
// Take care of the last value for odd n
if avalue < n*n {
avalue++
values[n/2][n/2] = avalue
}
// Ensure the same spacing for each element
places := int(n*n/10)
entry_width := 1
for places > 0 {
entry_width++
places /= 10
}
format := fmt.Sprintf("%%%dd ", entry_width)
// Print out the snail
for i := 0; i < n; i++ {
for j := 0; j < n ; j++ {
fmt.Printf(format, values[i][j])
}
fmt.Println("")
}
}
What about you? Have you tried solving this problem in golang? How would you do it better?
