Bottom up mergesort - singly linked list

I have written the following program in Go to mergesort a singly linked list. It does bottom up merge sort and hence does not have to find the mid of the linked list. Please review the algorithm and the code for idiomatic Go.

```
package main

import (
"fmt"
)

type node struct {
val int
next *node
}

type list struct {
head *node
}

func newList(arr []int) *list {

if arr == nil || len(arr) == 0 {
return nil
}
l := new(list)
l.head = getNode(arr[0])
prev := l.head
for i := 1; i < len(arr); i++ {
n := getNode(arr[i])
prev.next = n
prev = n
}

return l
}

func printList(n *node) {
for p := n; p != nil; p = p.next {
fmt.Printf("%d ", p.val)
}
}

func getNode(v int) *node {
n := new(node)
n.val = v
return n
}

func skip(n node, len int) (fast node, slow *node) {

counter := 0
fast = n
slow = n
for counter < len {
if fast != nil && fast.next != nil {
fast = fast.next.next
} else if fast != nil {
fast = fast.next
}

if slow != nil {
slow = slow.next
}

counter++
}

return fast, slow

}

func (l *list) mergesort() {

// start with sublength of 1
// compare 0,with 1, 2 with 3
// skip n.next.next for next bunch

sublen := 1
len := 0
counted := false

for !counted || sublen < len {
for n := l.head; n != nil; {
fast, slow := skip(n, sublen)

merge(n, slow, sublen)

if !counted {
len += 2
}
n = fast
}
counted = true
sublen += sublen
}
}

func merge(p node, q node, len int) {

i := p

if q == nil {
// nothing to merge, return
return
}

j := q
ilen := 0
jlen := 0

var head *node
var prev *node
var end *node
var penult *node

for (i != q && ilen < len) && (j != nil

Just some random notes:

l := new(list)
l.head = getNode(arr[0])

This is usually spelled l := &list{head: getNode(arr[0])}.

n := new(node)
n.val = v

Again, can be shortened to n := &node{val: v}.

var head *node
var prev *node
var end *node
var penult *node

Shorten to var head, prev, end, penult *node.

func append

This is a very bad name for a function because it clashes with a built-in. Better change to something like appendList or make it a method.

StackExchange Code Review Q#100519, answer score: 2