I am trying to learn Go, so here is my very simple function for removing adjacent duplicates from slice for exercise from the book by Donovan & Kernighan.
Here is the code: https://play.golang.org/p/avHc1ixfck
package main
import "fmt"
func main() {
a := []int{0, 1, 1, 3, 3, 3}
removeDup(a)
fmt.Println(a)
}
func removeDup(s []int) {
n := len(s)
tmp := make([]int, 0, n)
tmp = append(tmp, s[0])
j := 1
for i := 1; i < n; i++ {
if s[i] != s[i-1] {
tmp = append(tmp, s[i])
j++
}
}
s = s[:len(tmp)]
copy(s, tmp)
}
It should print out [0 1 3] - and I checked, actually tmp at the end of the function it has desired form. However, the result is [0 1 3 3 3 3]. I guess there is something with copy function.
Can I somehow replace input slice s with the temp or trim it to desired length?
Option 1
Return a new slice as suggested by #zerkms.
https://play.golang.org/p/uGJiD3WApS
package main
import "fmt"
func main() {
a := []int{0, 1, 1, 3, 3, 3}
a = removeDup(a)
fmt.Println(a)
}
func removeDup(s []int) []int {
n := len(s)
tmp := make([]int, 0, n)
tmp = append(tmp, s[0])
for i := 1; i < n; i++ {
if s[i] != s[i-1] {
tmp = append(tmp, s[i])
}
}
return tmp
}
Option 2
Use pointers for pass-by-reference.
The same thing in effect as that of option1.
https://play.golang.org/p/80bE5Qkuuj
package main
import "fmt"
func main() {
a := []int{0, 1, 1, 3, 3, 3}
removeDup(&a)
fmt.Println(a)
}
func removeDup(sp *[]int) {
s := *sp
n := len(s)
tmp := make([]int, 0, n)
tmp = append(tmp, s[0])
for i := 1; i < n; i++ {
if s[i] != s[i-1] {
tmp = append(tmp, s[i])
}
}
*sp = tmp
}
Also, refer to following SO thread:
Does Go have no real way to shrink a slice? Is that an issue?
Here's two more slightly different ways to achieve what you want using sets and named types. The cool thing about named types is that you can create interfaces around them and can help with the readability of lots of code.
package main
import "fmt"
func main() {
// returning a list
a := []int{0, 1, 1, 3, 3, 3}
clean := removeDup(a)
fmt.Println(clean)
// creating and using a named type
nA := &newArrType{0, 1, 1, 3, 3, 3}
nA.removeDup2()
fmt.Println(nA)
// or... casting your orginal array to the named type
nB := newArrType(a)
nB.removeDup2()
fmt.Println(nB)
}
// using a set
// order is not kept, but a set is returned
func removeDup(s []int) (newArr []int) {
set := make(map[int]struct{})
for _, n := range s {
set[n] = struct{}{}
}
newArr = make([]int, 0, len(set))
for k := range set {
newArr = append(newArr, k)
}
return
}
// using named a typed
type newArrType []int
func (a *newArrType) removeDup2() {
x := *a
for i := range x {
f := i + 1
if f < len(x) {
if x[i] == x[f] {
x = x[:f+copy(x[f:], x[f+1:])]
}
}
}
// check the last 2 indexes
if x[len(x)-2] == x[len(x)-1] {
x = x[:len(x)-1+copy(x[len(x)-1:], x[len(x)-1+1:])]
}
*a = x
}
Related
enter image description here
This code snippet is for give two slice of binary number a1 and a2 to return sum slice r1, and I want to figure out how long spend with this code snippet figure out the result.
and I figure out the factorial result.
Is my analysis right?
my analysis for time complexity is:
cn + (n*n!) + c
the Code is:
func BinaryPlus(a1 []int, a2 []int) []int {
var r1 = make([]int, len(a1), 2*(len(a1)))
for i := 0; i < len(a1); i++ {
r1[i] = a1[i] + a2[i]
}
// 二分反转
ReverseSlice(r1)
r1 = append(r1, 0)
final := 0
for i := 0; final != 1; i++ {
isOver := 1
for j := 0; j < len(r1); j++ {
if r1[j] > 1 {
r1[j] = r1[j] % 2
r1[j+1] += 1
if r1[j+1] > 1 {
isOver = 0
}
}
}
if isOver == 1 {
final = 1
}
}
// 二分反转
ReverseSlice(r1)
return r1
}
func ReverseSlice(s interface{}) {
n := reflect.ValueOf(s).Len()
swap := reflect.Swapper(s)
for i, j := 0, n-1; i < j; i, j = i+1, j-1 {
swap(i, j)
}
}
It is not entirely clear that your code, as written, is correct. The size cap for your result array could be too small. Consider the case that len(a2) > 2*len(a1): the r1 := make(...) will not reserve enough in this case. Further, the initial for loop will miss adding in the more significant bits of a2.
Binary addition should have no more than O(n) complexity. You can do it with a single for loop. n = 1+max(len(a1),len(a2)):
package main
import (
"fmt"
"reflect"
)
func BinaryPlus(a1 []int, a2 []int) []int {
reserve := len(a1) + 1
if x := len(a2) + 1; x > reserve {
reserve = x
}
hold := 0
maxBit := 1
ans := make([]int, reserve)
for i := 1; i <= reserve; i++ {
hold = hold / 2
if i <= len(a1) {
hold += a1[len(a1)-i]
}
if i <= len(a2) {
hold += a2[len(a2)-i]
}
ans[reserve-i] = hold & 1
if hold != 0 && i > maxBit {
maxBit = i
}
}
return ans[reserve-maxBit:]
}
func main() {
tests := []struct {
a, b, want []int
}{
{
a: []int{1},
b: []int{0},
want: []int{1},
},
{
a: []int{1, 0},
b: []int{0, 0, 1},
want: []int{1, 1},
},
{
a: []int{1, 0, 0, 1},
b: []int{1, 1, 1, 1, 0},
want: []int{1, 0, 0, 1, 1, 1},
},
{
a: []int{0, 0},
b: []int{0, 0, 0, 0, 0},
want: []int{0},
},
}
bad := false
for i := 0; i < len(tests); i++ {
t := tests[i]
c := BinaryPlus(t.a, t.b)
if !reflect.DeepEqual(c, t.want) {
fmt.Println(t.a, "+", t.b, "=", c, "; wanted:", t.want)
bad = true
}
}
if bad {
fmt.Println("FAILED")
} else {
fmt.Println("PASSED")
}
}
I can't get this Go lang test program to run. The compiler keeps giving an error on the append() function call below with an "evaluated but not used" error. I can't figure out why.
package main
import (
"fmt"
)
func removeDuplicates(testArr *[]int) int {
prevValue := (*testArr)[0]
for curIndex := 1; curIndex < len((*testArr)); curIndex++ {
curValue := (*testArr)[curIndex]
if curValue == prevValue {
append((*testArr)[:curIndex], (*testArr)[curIndex+1:]...)
}
prevValue = curValue
}
return len(*testArr)
}
func main() {
testArr := []int{0, 0, 1, 1, 1, 2, 2, 3, 3, 4}
nonDupSize := removeDuplicates(&testArr)
fmt.Printf("nonDupSize = %d", nonDupSize)
}
"evaluated but not used" error.
Code below is my idea. I think your code is not very clear.
package main
import (
"fmt"
)
func removeDuplicates(testArr *[]int) int {
m := make(map[int]bool)
arr := make([]int, 0)
for curIndex := 0; curIndex < len((*testArr)); curIndex++ {
curValue := (*testArr)[curIndex]
if has :=m[curValue]; !has {
m[curValue] = true
arr = append(arr, curValue)
}
}
*testArr = arr
return len(*testArr)
}
func main() {
testArr := []int{0, 0, 1, 1, 1, 2, 2, 3, 3, 4}
nonDupSize := removeDuplicates(&testArr)
fmt.Printf("nonDupSize = %d", nonDupSize)
}
Peter's answer nailed it, the compile error was due to not grabbing the return value from append()
Can anyone comment on whether this is a reasonable and idiomatic way of implementing circular shift of integer arrays in Go? (I deliberately chose not to use bitwise operations.)
How could it be improved?
package main
import "fmt"
func main() {
a := []int{1,2,3,4,5,6,7,8,9,10}
fmt.Println(a)
rotateR(a, 5)
fmt.Println(a)
rotateL(a, 5)
fmt.Println(a)
}
func rotateL(a []int, i int) {
for count := 1; count <= i; count++ {
tmp := a[0]
for n := 1;n < len(a);n++ {
a[n-1] = a[n]
}
a[len(a)-1] = tmp
}
}
func rotateR(a []int, i int) {
for count := 1; count <= i; count++ {
tmp := a[len(a)-1]
for n := len(a)-2;n >=0 ;n-- {
a[n+1] = a[n]
}
a[0] = tmp
}
}
Rotating the slice one position at a time, and repeating to get the total desired rotation means it will take time proportional to rotation distance × length of slice. By moving each element directly into its final position you can do this in time proportional to just the length of the slice.
The code for this is a little more tricky than you have, and you’ll need a GCD function to determine how many times to go through the slice:
func gcd(a, b int) int {
for b != 0 {
a, b = b, a % b
}
return a
}
func rotateL(a []int, i int) {
// Ensure the shift amount is less than the length of the array,
// and that it is positive.
i = i % len(a)
if i < 0 {
i += len(a)
}
for c := 0; c < gcd(i, len(a)); c++ {
t := a[c]
j := c
for {
k := j + i
// loop around if we go past the end of the slice
if k >= len(a) {
k -= len(a)
}
// end when we get to where we started
if k == c {
break
}
// move the element directly into its final position
a[j] = a[k]
j = k
}
a[j] = t
}
}
Rotating a slice of size l right by p positions is equivalent to rotating it left by l − p positions, so you can simplify your rotateR function by using rotateL:
func rotateR(a []int, i int) {
rotateL(a, len(a) - i)
}
Your code is fine for in-place modification.
Don't clearly understand what you mean by bitwise operations. Maybe this
package main
import "fmt"
func main() {
a := []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
fmt.Println(a)
rotateR(&a, 4)
fmt.Println(a)
rotateL(&a, 4)
fmt.Println(a)
}
func rotateL(a *[]int, i int) {
x, b := (*a)[:i], (*a)[i:]
*a = append(b, x...)
}
func rotateR(a *[]int, i int) {
x, b := (*a)[:(len(*a)-i)], (*a)[(len(*a)-i):]
*a = append(b, x...)
}
Code works https://play.golang.org/p/0VtiRFQVl7
It's called reslicing in Go vocabulary. Tradeoff is coping and looping in your snippet vs dynamic allocation in this. It's your choice, but in case of shifting 10000 elements array by one position reslicing looks much cheaper.
I like Uvelichitel solution but if you would like modular arithmetic which would be O(n) complexity
package main
func main(){
s := []string{"1", "2", "3"}
rot := 5
fmt.Println("Before RotL", s)
fmt.Println("After RotL", rotL(rot, s))
fmt.Println("Before RotR", s)
fmt.Println("After RotR", rotR(rot,s))
}
func rotL(m int, arr []string) []string{
newArr := make([]string, len(arr))
for i, k := range arr{
newPos := (((i - m) % len(arr)) + len(arr)) % len(arr)
newArr[newPos] = k
}
return newArr
}
func rotR(m int, arr []string) []string{
newArr := make([]string, len(arr))
for i, k := range arr{
newPos := (i + m) % len(arr)
newArr[newPos] = k
}
return newArr
}
If you need to enter multiple values, whatever you want (upd code Uvelichitel)
package main
import "fmt"
func main() {
var N, n int
fmt.Scan(&N)
a := make([]int, N)
for i := 0; i < N; i++ {
fmt.Scan(&a[i])
}
fmt.Scan(&n)
if n > 0 {
rotateR(&a, n%len(a))
} else {
rotateL(&a, (n*-1)%len(a))
}
for _, elem := range a {
fmt.Print(elem, " ")
}
}
func rotateL(a *[]int, i int) {
x, b := (*a)[:i], (*a)[i:]
*a = append(b, x...)
}
func rotateR(a *[]int, i int) {
x, b := (*a)[:(len(*a)-i)], (*a)[(len(*a)-i):]
*a = append(b, x...)
}
How do I reverse an arbitrary slice ([]interface{}) in Go? I'd rather not have to write Less and Swap to use sort.Reverse. Is there a simple, builtin way to do this?
The standard library does not have a built-in function for reversing a slice. Use a for loop to reverse a slice:
for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
s[i], s[j] = s[j], s[i]
}
Use type parameters to write a generic reverse function in Go 1.18 or later:
func reverse[S ~[]E, E any](s S) {
for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
s[i], s[j] = s[j], s[i]
}
}
Use reflect.Swapper to write a function that works with arbitrary slice types in Go version 1.8 or later:
func reverse(s interface{}) {
n := reflect.ValueOf(s).Len()
swap := reflect.Swapper(s)
for i, j := 0, n-1; i < j; i, j = i+1, j-1 {
swap(i, j)
}
}
Run the code on the Go playground.
The functions in this answer reverse the slice inplace. If you do not want to modify the original slice, copy the slice before reversing the slice.
Here's another possible way to reverse generic slice (go 1.18)
// You can edit this code!
// Click here and start typing.
package main
import (
"fmt"
"sort"
)
func main() {
nums := []int64{10, 5, 15, 20, 1, 100, -1}
ReverseSlice(nums)
fmt.Println(nums)
strs := []string{"hello", "world"}
ReverseSlice(strs)
fmt.Println(strs)
runes := []rune{'h', 'e', 'l', 'l', 'o', 'w', 'o', 'r', 'l', 'd'}
ReverseSlice(runes)
for _, r := range runes {
fmt.Print(string(r), " ")
}
}
func ReverseSlice[T comparable](s []T) {
sort.SliceStable(s, func(i, j int) bool {
return i > j
})
}
Running the program above should output:
[-1 100 1 20 15 5 10]
[world hello]
d l r o w o l l e h
Program exited.
go playground
This will return a reversed slice without modifying the original slice.
Algorithm used from official wiki page: https://github.com/golang/go/wiki/SliceTricks#reversing
func reverse(s []interface{}) []interface{} {
a := make([]interface{}, len(s))
copy(a, s)
for i := len(a)/2 - 1; i >= 0; i-- {
opp := len(a) - 1 - i
a[i], a[opp] = a[opp], a[i]
}
return a
}
There are my code example, you can run it in playground
package main
import (
"fmt"
"reflect"
"errors"
)
func ReverseSlice(data interface{}) {
value := reflect.ValueOf(data)
if value.Kind() != reflect.Slice {
panic(errors.New("data must be a slice type"))
}
valueLen := value.Len()
for i := 0; i <= int((valueLen-1)/2); i++ {
reverseIndex := valueLen - 1 - i
tmp := value.Index(reverseIndex).Interface()
value.Index(reverseIndex).Set(value.Index(i))
value.Index(i).Set(reflect.ValueOf(tmp))
}
}
func main() {
names := []string{"bob", "mary", "sally", "michael"}
ReverseSlice(names)
fmt.Println(names)
}
Here is the function I'm using with generics (go 1.18+). You can use it to reverse any kind of slice or even a string (using the split/join trick). It doesn't change the original slice.
package main
import (
"fmt"
"strings"
)
func Reverse[T any](original []T) (reversed []T) {
reversed = make([]T, len(original))
copy(reversed, original)
for i := len(reversed)/2 - 1; i >= 0; i-- {
tmp := len(reversed) - 1 - i
reversed[i], reversed[tmp] = reversed[tmp], reversed[i]
}
return
}
func main() {
a := []string{"a", "b", "c"}
fmt.Println(a, Reverse(a))
b := []uint{0, 1, 2}
fmt.Println(b, Reverse(b))
c := "abc"
fmt.Println(c, strings.Join(Reverse(strings.Split(c, "")), ""))
}
Better Go Playground
This generic slice reversal function should do it for you:
func ReverseSlice[T comparable](s []T) []T {
var r []T
for i := len(s) - 1; i >= 0; i-- {
r = append(r, s[i])
}
return r
}
It would be convenient to be able to say something like:
for _, element := reverse range mySlice {
...
}
Edit: I asked this question a long time ago, it is 2022 now and the generic solution by #Ivan below seems like the way to go!
No there is no convenient operator for this to add to the range one in place. You'll have to do a normal for loop counting down:
s := []int{5, 4, 3, 2, 1}
for i := len(s)-1; i >= 0; i-- {
fmt.Println(s[i])
}
You can also do:
s := []int{5, 4, 3, 2, 1}
for i := range s {
fmt.Println(s[len(s)-1-i]) // Suggestion: do `last := len(s)-1` before the loop
}
Output:
1
2
3
4
5
Also here: http://play.golang.org/p/l7Z69TV7Vl
Variation with index
for k := range s {
k = len(s) - 1 - k
// now k starts from the end
}
How about use defer:
s := []int{5, 4, 3, 2, 1}
for i, _ := range s {
defer fmt.Println(s[i])
}
One could use a channel to reverse a list in a function without duplicating it. It makes the code nicer in my sense.
package main
import (
"fmt"
)
func reverse(lst []string) chan string {
ret := make(chan string)
go func() {
for i, _ := range lst {
ret <- lst[len(lst)-1-i]
}
close(ret)
}()
return ret
}
func main() {
elms := []string{"a", "b", "c", "d"}
for e := range reverse(elms) {
fmt.Println(e)
}
}
In 2022, you could use generics to reverse any slice in-place:
func reverse[S ~[]E, E any](s S) {
for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
s[i], s[j] = s[j], s[i]
}
}
When I need to extract elements from a slice and reverse range, I use something like this code:
// reverse range
// Go Playground: https://play.golang.org/p/gx6fJIfb7fo
package main
import (
"fmt"
)
type Elem struct {
Id int64
Name string
}
type Elems []Elem
func main() {
mySlice := Elems{{Id: 0, Name: "Alice"}, {Id: 1, Name: "Bob"}, {Id: 2, Name: "Carol"}}
for i, element := range mySlice {
fmt.Printf("Normal range: [%v] %+v\n", i, element)
}
//mySlice = Elems{}
//mySlice = Elems{{Id: 0, Name: "Alice"}}
if last := len(mySlice) - 1; last >= 0 {
for i, element := last, mySlice[0]; i >= 0; i-- {
element = mySlice[i]
fmt.Printf("Reverse range: [%v] %+v\n", i, element)
}
} else {
fmt.Println("mySlice empty")
}
}
Output:
Normal range: [0] {Id:0 Name:Alice}
Normal range: [1] {Id:1 Name:Bob}
Normal range: [2] {Id:2 Name:Carol}
Reverse range: [2] {Id:2 Name:Carol}
Reverse range: [1] {Id:1 Name:Bob}
Reverse range: [0] {Id:0 Name:Alice}
Playground: https://play.golang.org/p/gx6fJIfb7fo
You can use the funk.ForEachRight method from go-funk:
results := []int{}
funk.ForEachRight([]int{1, 2, 3, 4}, func(x int) {
results = append(results, x)
})
fmt.Println(results) // []int{4, 3, 2, 1}