Finding longest word in golang - string

Trying to find the longest word using Go from a sentence.
At the moment I am using this method:
func longestWord(s string) string {
newArr := strings.Split(s, " ")
l := len(newArr[0])
long := newArr[0]
var result string
// fmt.Println(long)
for _, lenString := range newArr {
if len(lenString) > l {
// ll := len(lenString)
// l := len(lenString)
d := &l
p := &long
c := &result
*d = len(lenString)
*p = lenString
*c = lenString
// fmt.Println(lenString)
} else {
c := &result
*c = newArr[0]
}
}
return result
}
func main() {
args := "Monday Tuesday Friday Sunday Wednesday"
fmt.Println(longestWord(args))
}
But I'm not sure that this is the best method to achieve that. Is there any other elegant way to do that? I know that there is one more method by using sort, but I would prefer more using the way with iteration between words.

"Best" solution
We can even write it more compact than the other answers by taking advantage of the following:
using tuple assignments
initializing the best and its length with the zero values ("" and 0) and omitting the check for 0 words as the for range handles that properly
no need to store words as a local variable as it is only used in the loop
We lose nothing from readability:
func longestWord(s string) string {
best, length := "", 0
for _, word := range strings.Split(s, " ") {
if len(word) > length {
best, length = word, len(word)
}
}
return best
}
Testing it:
fmt.Printf("%q\n", longestWord(""))
args := "Monday Tuesday Friday Sunday Wednesday"
fmt.Printf("%q\n", longestWord(args))
Output (try it on the Go Playground):
""
"Wednesday"
Most compact solution
Note that storing the length of the best is optional and is purely for optimization purposes, since if we have best, its length is always len(best).
Taking advantage of this, and that we can use named result parameters (and that all variables are initialized to the zero value of their types unless an initial value is provided–which for string is ""), we can even write it more compact, again losing nothing from readability:
func longestWord(s string) (best string) {
for _, word := range strings.Split(s, " ") {
if len(word) > len(best) {
best = word
}
}
return
}
Testing and output is the same, try it on the Go Playground. Again, in most cases this is probably slightly slower compared to when we stored the length too.

That totally works! You could make it a bit shorter, while also using longer variable names that explain a bit more about your intention.
func longestWord(s string) string {
words := strings.Split(s, " ")
if len(words) == 0 {
return ""
}
best := words[0]
best_length := 0
for _, word := range words {
if len(word) > best_length {
best = word
best_length = len(word)
}
}
return best
}
You could change this to track a pointer instead of the word itself if you like.

I would do it like this:
func longestWord(s string) string {
newArr := strings.Split(s, " ")
longestWord := ""
longestLength := 0
// loop through the array
for _, word := range newArr {
// save length of word in the actual iteration
length := len(word)
// if length is larger, than longest
if length > longestLength {
// save the new longest word
longestWord = word
longestLength = length
}
}
// return the longest word
return longestWord
}
Implementation can be found on the go playground

Related

How to print each elements string with its frequency based on index in map - Golang

Let say I have input and output string where the output will be the frequency of each elements in the string and the char itself
input := "programming"
output := "p2ro2ga2min"
How can I print it based on index after I found the freq of distinct character
This is my code
func countFreq(s string) {
sMap := make(map[string]int)
for _, v := range s {
sMap[string(v)]++
}
for i, v := range sMap {
fmt.Printf("%v%v", i, v)
}
// Don't know what to do next
}
The output of code is
output: n1p1r2o1g2a1m2i1
#icza's answer is great. Here's an alternative that I thought of before but only just got around to writing.
It uses a string to keep track of the rune order but you could use a string builder if speed is important.
func countFreq(s string) {
sMap := make(map[rune]int)
sOut := ""
for _, c := range s {
sMap[c]++
if sMap[c] == 1 {
sOut += string(c)
}
}
for _, c := range sOut {
if sMap[c] > 1 {
fmt.Print(sMap[c])
}
fmt.Printf("%c", c)
}
}
You're counting runes, so use a map of map[rune]int, so you can omit the conversions back to string.
Maps are unordered, so if you want the output in the same order as the input, you can't (shouldn't) iterate over the map.
Once you counted the letters, range over the input letters again, and get the frequency from the map, and remove it. If the count is greater than 1, also print the number.
func countFreq(s string) {
sMap := make(map[rune]int)
for _, v := range s {
sMap[v]++
}
for _, v := range s {
count := sMap[v]
if count == 0 {
continue // Char already printed and removed
}
delete(sMap, v)
if count > 1 {
fmt.Print(count)
}
fmt.Print(string(v))
}
}
Testing it:
for _, s := range []string{"programming", "ab", "aba", "aabcdeac"} {
fmt.Println("In:", s)
fmt.Print("Out: ")
countFreq(s)
fmt.Println()
}
This will output (try it on the Go Playground):
In: programming
Out: p2ro2ga2min
In: ab
Out: ab
In: aba
Out: 2ab
In: aabcdeac
Out: 3ab2cde

Splitting a rune correctly in golang

I'm wondering if there is an easy way, such as well known functions to handle code points/runes, to take a chunk out of the middle of a rune slice without messing it up or if it's all needs to coded ourselves to get down to something equal to or less than a maximum number of bytes.
Specifically, what I am looking to do is pass a string to a function, convert it to runes so that I can respect code points and if the slice is longer than some maximum bytes, remove enough runes from the center of the runes to get the bytes down to what's necessary.
This is simple math if the strings are just single byte characters and be handled something like:
func shortenStringIDToMaxLength(in string, maxLen int) string {
if len(in) > maxLen {
excess := len(in) - maxLen
start := maxLen/2 - excess/2
return in[:start] + in[start+excess:]
}
return in
}
but in a variable character width byte string it's either going to be a fair bit more coding looping through or there will be nice functions to make this easy. Does anyone have a code sample of how to best handle such a thing with runes?
The idea here is that the DB field the string will go into has a fixed maximum length in bytes, not code points so there needs to be some algorithm from runes to maximum bytes. The reason for taking the characters from the the middle of the string is just the needs of this particular program.
Thanks!
EDIT:
Once I found out that the range operator respected runes on strings this became easy to do with just strings which I found because of the great answers below. I shouldn't have to worry about the string being a well formed UTF format in this case but if I do I now know about the UTF module, thanks!
Here's what I ended up with:
package main
import (
"fmt"
)
func ShortenStringIDToMaxLength(in string, maxLen int) string {
if maxLen < 1 {
// Panic/log whatever is your error system of choice.
}
bytes := len(in)
if bytes > maxLen {
excess := bytes - maxLen
lPos := bytes/2 - excess/2
lastPos := 0
for pos, _ := range in {
if pos > lPos {
lPos = lastPos
break
}
lastPos = pos
}
rPos := lPos + excess
for pos, _ := range in[lPos:] {
if pos >= excess {
rPos = pos
break
}
}
return in[:lPos] + in[lPos+rPos:]
}
return in
}
func main() {
out := ShortenStringIDToMaxLength(`123456789 123456789`, 5)
fmt.Println(out, len(out))
}
https://play.golang.org/p/YLGlj_17A-j
Here is an adaptation of your algorithm, which removes incomplete runes from the beginning of your prefix and the end of your suffix :
func TrimLastIncompleteRune(s string) string {
l := len(s)
for i := 1; i <= l; i++ {
suff := s[l-i : l]
// repeatedly try to decode a rune from the last bytes in string
r, cnt := utf8.DecodeRuneInString(suff)
if r == utf8.RuneError {
continue
}
// if success : return the substring which contains
// this succesfully decoded rune
lgth := l - i + cnt
return s[:lgth]
}
return ""
}
func TrimFirstIncompleteRune(s string) string {
// repeatedly try to decode a rune from the beginning
for i := 0; i < len(s); i++ {
if r, _ := utf8.DecodeRuneInString(s[i:]); r != utf8.RuneError {
// if success : return
return s[i:]
}
}
return ""
}
func shortenStringIDToMaxLength(in string, maxLen int) string {
if len(in) > maxLen {
firstHalf := maxLen / 2
secondHalf := len(in) - (maxLen - firstHalf)
prefix := TrimLastIncompleteRune(in[:firstHalf])
suffix := TrimFirstIncompleteRune(in[secondHalf:])
return prefix + suffix
}
return in
}
link on play.golang.org
This algorithm only tries to drop more bytes from the selected prefix and suffix.
If it turns out that you need to drop 3 bytes from the suffix to have a valid rune, for example, it does not try to see if it can add 3 more bytes to the prefix, to have an end result closer to maxLen bytes.
You can use simple arithmetic to find start and end such that the string s[:start] + s[end:] is shorter than your byte limit. But you need to make sure that start and end are both the first byte of any utf-8 sequence to keep the sequence valid.
UTF-8 has the property that any given byte is the first byte of a sequence as long as its top two bits aren't 10.
So you can write code something like this (playground: https://play.golang.org/p/xk_Yo_1wTYc)
package main
import (
"fmt"
)
func truncString(s string, maxLen int) string {
if len(s) <= maxLen {
return s
}
start := (maxLen + 1) / 2
for start > 0 && s[start]>>6 == 0b10 {
start--
}
end := len(s) - (maxLen - start)
for end < len(s) && s[end]>>6 == 0b10 {
end++
}
return s[:start] + s[end:]
}
func main() {
fmt.Println(truncString("this is a test", 5))
fmt.Println(truncString("日本語", 7))
}
This code has the desirable property that it takes O(maxLen) time, no matter how long the input string (assuming it's valid utf-8).

Expand a string of slice by delimiter

I want to expand a string of slice by delimiter "/".
For example, expanding the following slice
s := []string{"5/3","9","5/4/1","6"}
Should produce individual slices :
["5","9","5","6"] ["5","9","4","6"] ["5","9","1","6"]
["3","9","5","6"] ["3","9","4","6"] ["3","9","1","6"]
I am pretty much stuck here
var c [][]string{}
s := []string{"5/3","9","5/4/1","6"}
for _, v := range s {
combos := strings.Split(v, "/")
for _, combo := range combos {
}
}
Running time aside, you can achieve this with recursion.
func Perm(digits [][]string) (perm [][]string) {
if len(digits) == 0 || len(digits) == 1 {
return digits
}
nextDigits := Perm(digits[1:])
for _, digit := range digits[0] {
for _, next := range nextDigits {
cat := append([]string{digit}, next...)
perm = append(perm, cat)
}
}
return perm
}
Playground

Bitmasking conversion of CPU ids with Go

I have a mask that contains a binary counting of cpu_ids (0xA00000800000 for 3 CPUs) which I want to convert into a string of comma separated cpu_ids: "0,2,24".
I did the following Go implementation (I am a Go starter). Is it the best way to do it? Especially the handling of byte buffers seems to be inefficient!
package main
import (
"fmt"
"os"
"os/exec"
)
func main(){
cpuMap := "0xA00000800000"
cpuIds = getCpuIds(cpuMap)
fmt.Println(cpuIds)
}
func getCpuIds(cpuMap string) string {
// getting the cpu ids
cpu_ids_i, _ := strconv.ParseInt(cpuMap, 0, 64) // int from string
cpu_ids_b := strconv.FormatInt(cpu_ids_i, 2) // binary as string
var buff bytes.Buffer
for i, runeValue := range cpu_ids_b {
// take care! go returns code points and not the string
if runeValue == '1' {
//fmt.Println(bitString, i)
buff.WriteString(fmt.Sprintf("%d", i))
}
if (i+1 < len(cpu_ids_b)) && (runeValue == '1') {
//fmt.Println(bitString)
buff.WriteString(string(","))
}
}
cpuIds := buff.String()
// remove last comma
cpuIds = cpuIds[:len(cpuIds)-1]
//fmt.Println(cpuIds)
return cpuIds
}
Returns:
"0,2,24"
What you're doing is essentially outputting the indices of the "1"'s in the binary representation from left-to-right, and starting index counting from the left (unusal).
You can achieve the same using bitmasks and bitwise operators, without converting it to a binary string. And I would return a slice of indices instead of its formatted string, easier to work with.
To test if the lowest (rightmost) bit is 1, you can do it like x&0x01 == 1, and to shift a whole number bitwise to the right: x >>= 1. After a shift, the rightmost bit "disappears", and the previously 2nd bit becomes the 1st, so you can test again with the same logic. You may loop until the number is greater than 0 (which means it sill has 1-bits).
See this question for more examples of bitwise operations: Difference between some operators "|", "^", "&", "&^". Golang
Of course if we test the rightmost bit and shift right, we get the bits (indices) in reverse order (compared to what you want), and the indices are counted from right, so we have to correct this before returning the result.
So the solution looks like this:
func getCpuIds(cpuMap string) (r []int) {
ci, err := strconv.ParseInt(cpuMap, 0, 64)
if err != nil {
panic(err)
}
count := 0
for ; ci > 0; count, ci = count+1, ci>>1 {
if ci&0x01 == 1 {
r = append(r, count)
}
}
// Indices are from the right, correct it:
for i, v := range r {
r[i] = count - v - 1
}
// Result is in reverse order:
for i, j := 0, len(r)-1; i < j; i, j = i+1, j-1 {
r[i], r[j] = r[j], r[i]
}
return
}
Output (try it on the Go Playground):
[0 2 24]
If for some reason you need the result as a comma separated string, this is how you can obtain that:
buf := &bytes.Buffer{}
for i, v := range cpuIds {
if i > 0 {
buf.WriteString(",")
}
buf.WriteString(strconv.Itoa(v))
}
cpuIdsStr := buf.String()
fmt.Println(cpuIdsStr)
Output (try it on the Go Playground):
0,2,24

How to reverse a string in Go?

How can we reverse a simple string in Go?
In Go1 rune is a builtin type.
func Reverse(s string) string {
runes := []rune(s)
for i, j := 0, len(runes)-1; i < j; i, j = i+1, j-1 {
runes[i], runes[j] = runes[j], runes[i]
}
return string(runes)
}
Russ Cox, on the golang-nuts mailing list, suggests
package main
import "fmt"
func main() {
input := "The quick brown 狐 jumped over the lazy 犬"
// Get Unicode code points.
n := 0
rune := make([]rune, len(input))
for _, r := range input {
rune[n] = r
n++
}
rune = rune[0:n]
// Reverse
for i := 0; i < n/2; i++ {
rune[i], rune[n-1-i] = rune[n-1-i], rune[i]
}
// Convert back to UTF-8.
output := string(rune)
fmt.Println(output)
}
This works, without all the mucking about with functions:
func Reverse(s string) (result string) {
for _,v := range s {
result = string(v) + result
}
return
}
From Go example projects: golang/example/stringutil/reverse.go, by Andrew Gerrand
/*
Copyright 2014 Google Inc.
Licensed under the Apache License, Version 2.0 (the "License");
you may not use this file except in compliance with the License.
You may obtain a copy of the License at
http://www.apache.org/licenses/LICENSE-2.0
Unless required by applicable law or agreed to in writing, software
distributed under the License is distributed on an "AS IS" BASIS,
WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
See the License for the specific language governing permissions and
limitations under the License.
*/
// Reverse returns its argument string reversed rune-wise left to right.
func Reverse(s string) string {
r := []rune(s)
for i, j := 0, len(r)-1; i < len(r)/2; i, j = i+1, j-1 {
r[i], r[j] = r[j], r[i]
}
return string(r)
}
Go Playground for reverse a string
After reversing string "bròwn", the correct result should be "nwòrb", not "nẁorb".
Note the grave above the letter o.
For preserving Unicode combining characters such as "as⃝df̅" with reverse result "f̅ds⃝a",
please refer to another code listed below:
http://rosettacode.org/wiki/Reverse_a_string#Go
This works on unicode strings by considering 2 things:
range works on string by enumerating unicode characters
string can be constructed from int slices where each element is a unicode character.
So here it goes:
func reverse(s string) string {
o := make([]int, utf8.RuneCountInString(s));
i := len(o);
for _, c := range s {
i--;
o[i] = c;
}
return string(o);
}
There are too many answers here. Some of them are clear duplicates. But even from the left one, it is hard to select the best solution.
So I went through the answers, thrown away the one that does not work for unicode and also removed duplicates. I benchmarked the survivors to find the fastest. So here are the results with attribution (if you notice the answers that I missed, but worth adding, feel free to modify the benchmark):
Benchmark_rmuller-4 100000 19246 ns/op
Benchmark_peterSO-4 50000 28068 ns/op
Benchmark_russ-4 50000 30007 ns/op
Benchmark_ivan-4 50000 33694 ns/op
Benchmark_yazu-4 50000 33372 ns/op
Benchmark_yuku-4 50000 37556 ns/op
Benchmark_simon-4 3000 426201 ns/op
So here is the fastest method by rmuller:
func Reverse(s string) string {
size := len(s)
buf := make([]byte, size)
for start := 0; start < size; {
r, n := utf8.DecodeRuneInString(s[start:])
start += n
utf8.EncodeRune(buf[size-start:], r)
}
return string(buf)
}
For some reason I can't add a benchmark, so you can copy it from PlayGround (you can't run tests there). Rename it and run go test -bench=.
I noticed this question when Simon posted his solution which, since strings are immutable, is very inefficient. The other proposed solutions are also flawed; they don't work or they are inefficient.
Here's an efficient solution that works, except when the string is not valid UTF-8 or the string contains combining characters.
package main
import "fmt"
func Reverse(s string) string {
n := len(s)
runes := make([]rune, n)
for _, rune := range s {
n--
runes[n] = rune
}
return string(runes[n:])
}
func main() {
fmt.Println(Reverse(Reverse("Hello, 世界")))
fmt.Println(Reverse(Reverse("The quick brown 狐 jumped over the lazy 犬")))
}
I wrote the following Reverse function which respects UTF8 encoding and combined characters:
// Reverse reverses the input while respecting UTF8 encoding and combined characters
func Reverse(text string) string {
textRunes := []rune(text)
textRunesLength := len(textRunes)
if textRunesLength <= 1 {
return text
}
i, j := 0, 0
for i < textRunesLength && j < textRunesLength {
j = i + 1
for j < textRunesLength && isMark(textRunes[j]) {
j++
}
if isMark(textRunes[j-1]) {
// Reverses Combined Characters
reverse(textRunes[i:j], j-i)
}
i = j
}
// Reverses the entire array
reverse(textRunes, textRunesLength)
return string(textRunes)
}
func reverse(runes []rune, length int) {
for i, j := 0, length-1; i < length/2; i, j = i+1, j-1 {
runes[i], runes[j] = runes[j], runes[i]
}
}
// isMark determines whether the rune is a marker
func isMark(r rune) bool {
return unicode.Is(unicode.Mn, r) || unicode.Is(unicode.Me, r) || unicode.Is(unicode.Mc, r)
}
I did my best to make it as efficient and readable as possible. The idea is simple, traverse through the runes looking for combined characters then reverse the combined characters' runes in-place. Once we have covered them all, reverse the runes of the entire string also in-place.
Say we would like to reverse this string bròwn. The ò is represented by two runes, one for the o and one for this unicode \u0301a that represents the "grave".
For simplicity, let's represent the string like this bro'wn. The first thing we do is look for combined characters and reverse them. So now we have the string br'own. Finally, we reverse the entire string and end up with nwo'rb. This is returned to us as nwòrb
You can find it here https://github.com/shomali11/util if you would like to use it.
Here are some test cases to show a couple of different scenarios:
func TestReverse(t *testing.T) {
assert.Equal(t, Reverse(""), "")
assert.Equal(t, Reverse("X"), "X")
assert.Equal(t, Reverse("b\u0301"), "b\u0301")
assert.Equal(t, Reverse("😎⚽"), "⚽😎")
assert.Equal(t, Reverse("Les Mise\u0301rables"), "selbare\u0301siM seL")
assert.Equal(t, Reverse("ab\u0301cde"), "edcb\u0301a")
assert.Equal(t, Reverse("This `\xc5` is an invalid UTF8 character"), "retcarahc 8FTU dilavni na si `�` sihT")
assert.Equal(t, Reverse("The quick bròwn 狐 jumped over the lazy 犬"), "犬 yzal eht revo depmuj 狐 nwòrb kciuq ehT")
}
//Reverse reverses string using strings.Builder. It's about 3 times faster
//than the one with using a string concatenation
func Reverse(in string) string {
var sb strings.Builder
runes := []rune(in)
for i := len(runes) - 1; 0 <= i; i-- {
sb.WriteRune(runes[i])
}
return sb.String()
}
//Reverse reverses string using string
func Reverse(in string) (out string) {
for _, r := range in {
out = string(r) + out
}
return
}
BenchmarkReverseStringConcatenation-8 1000000 1571 ns/op 176 B/op 29 allocs/op
BenchmarkReverseStringsBuilder-8 3000000 499 ns/op 56 B/op 6 allocs/op
Using strings.Builder is about 3 times faster than using string concatenation
Here is quite different, I would say more functional approach, not listed among other answers:
func reverse(s string) (ret string) {
for _, v := range s {
defer func(r rune) { ret += string(r) }(v)
}
return
}
This is the fastest implementation
func Reverse(s string) string {
size := len(s)
buf := make([]byte, size)
for start := 0; start < size; {
r, n := utf8.DecodeRuneInString(s[start:])
start += n
utf8.EncodeRune(buf[size-start:], r)
}
return string(buf)
}
const (
s = "The quick brown 狐 jumped over the lazy 犬"
reverse = "犬 yzal eht revo depmuj 狐 nworb kciuq ehT"
)
func TestReverse(t *testing.T) {
if Reverse(s) != reverse {
t.Error(s)
}
}
func BenchmarkReverse(b *testing.B) {
for i := 0; i < b.N; i++ {
Reverse(s)
}
}
A simple stroke with rune:
func ReverseString(s string) string {
runes := []rune(s)
size := len(runes)
for i := 0; i < size/2; i++ {
runes[size-i-1], runes[i] = runes[i], runes[size-i-1]
}
return string(runes)
}
func main() {
fmt.Println(ReverseString("Abcdefg 汉语 The God"))
}
: doG ehT 语汉 gfedcbA
You could also import an existing implementation:
import "4d63.com/strrev"
Then:
strrev.Reverse("abåd") // returns "dåba"
Or to reverse a string including unicode combining characters:
strrev.ReverseCombining("abc\u0301\u031dd") // returns "d\u0301\u031dcba"
These implementations supports correct ordering of unicode multibyte and combing characters when reversed.
Note: Built-in string reverse functions in many programming languages do not preserve combining, and identifying combining characters requires significantly more execution time.
func ReverseString(str string) string {
output :=""
for _, char := range str {
output = string(char) + output
}
return output
}
// "Luizpa" -> "apziuL"
// "123日本語" -> "語本日321"
// "⚽😎" -> "😎⚽"
// "´a´b´c´" -> "´c´b´a´"
This code preserves sequences of combining characters intact, and
should work with invalid UTF-8 input too.
package stringutil
import "code.google.com/p/go.text/unicode/norm"
func Reverse(s string) string {
bound := make([]int, 0, len(s) + 1)
var iter norm.Iter
iter.InitString(norm.NFD, s)
bound = append(bound, 0)
for !iter.Done() {
iter.Next()
bound = append(bound, iter.Pos())
}
bound = append(bound, len(s))
out := make([]byte, 0, len(s))
for i := len(bound) - 2; i >= 0; i-- {
out = append(out, s[bound[i]:bound[i+1]]...)
}
return string(out)
}
It could be a little more efficient if the unicode/norm primitives
allowed iterating through the boundaries of a string without
allocating. See also https://code.google.com/p/go/issues/detail?id=9055 .
If you need to handle grapheme clusters, use unicode or regexp module.
package main
import (
"unicode"
"regexp"
)
func main() {
str := "\u0308" + "a\u0308" + "o\u0308" + "u\u0308"
println("u\u0308" + "o\u0308" + "a\u0308" + "\u0308" == ReverseGrapheme(str))
println("u\u0308" + "o\u0308" + "a\u0308" + "\u0308" == ReverseGrapheme2(str))
}
func ReverseGrapheme(str string) string {
buf := []rune("")
checked := false
index := 0
ret := ""
for _, c := range str {
if !unicode.Is(unicode.M, c) {
if len(buf) > 0 {
ret = string(buf) + ret
}
buf = buf[:0]
buf = append(buf, c)
if checked == false {
checked = true
}
} else if checked == false {
ret = string(append([]rune(""), c)) + ret
} else {
buf = append(buf, c)
}
index += 1
}
return string(buf) + ret
}
func ReverseGrapheme2(str string) string {
re := regexp.MustCompile("\\PM\\pM*|.")
slice := re.FindAllString(str, -1)
length := len(slice)
ret := ""
for i := 0; i < length; i += 1 {
ret += slice[length-1-i]
}
return ret
}
It's assuredly not the most memory efficient solution, but for a "simple" UTF-8 safe solution the following will get the job done and not break runes.
It's in my opinion the most readable and understandable on the page.
func reverseStr(str string) (out string) {
for _, s := range str {
out = string(s) + out
}
return
}
The following two methods run faster than the fastest solution that preserve combining characters, though that's not to say I'm missing something in my benchmark setup.
//input string s
bs := []byte(s)
var rs string
for len(bs) > 0 {
r, size := utf8.DecodeLastRune(bs)
rs += fmt.Sprintf("%c", r)
bs = bs[:len(bs)-size]
} // rs has reversed string
Second method inspired by this
//input string s
bs := []byte(s)
cs := make([]byte, len(bs))
b1 := 0
for len(bs) > 0 {
r, size := utf8.DecodeLastRune(bs)
d := make([]byte, size)
_ = utf8.EncodeRune(d, r)
b1 += copy(cs[b1:], d)
bs = bs[:len(bs) - size]
} // cs has reversed bytes
NOTE: This answer is from 2009, so there are probably better solutions out there by now.
Looks a bit 'roundabout', and probably not very efficient, but illustrates how the Reader interface can be used to read from strings. IntVectors also seem very suitable as buffers when working with utf8 strings.
It would be even shorter when leaving out the 'size' part, and insertion into the vector by Insert, but I guess that would be less efficient, as the whole vector then needs to be pushed back by one each time a new rune is added.
This solution definitely works with utf8 characters.
package main
import "container/vector";
import "fmt";
import "utf8";
import "bytes";
import "bufio";
func
main() {
toReverse := "Smørrebrød";
fmt.Println(toReverse);
fmt.Println(reverse(toReverse));
}
func
reverse(str string) string {
size := utf8.RuneCountInString(str);
output := vector.NewIntVector(size);
input := bufio.NewReader(bytes.NewBufferString(str));
for i := 1; i <= size; i++ {
rune, _, _ := input.ReadRune();
output.Set(size - i, rune);
}
return string(output.Data());
}
func Reverse(s string) string {
r := []rune(s)
var output strings.Builder
for i := len(r) - 1; i >= 0; i-- {
output.WriteString(string(r[i]))
}
return output.String()
}
Simple, Sweet and Performant
func reverseStr(str string) string {
strSlice := []rune(str) //converting to slice of runes
length := len(strSlice)
for i := 0; i < (length / 2); i++ {
strSlice[i], strSlice[length-i-1] = strSlice[length-i-1], strSlice[i]
}
return string(strSlice) //converting back to string
}
Reversing a string by word is a similar process. First, we convert the string into an array of strings where each entry is a word. Next, we apply the normal reverse loop to that array. Finally, we smush the results back together into a string that we can return to the caller.
package main
import (
"fmt"
"strings"
)
func reverse_words(s string) string {
words := strings.Fields(s)
for i, j := 0, len(words)-1; i < j; i, j = i+1, j-1 {
words[i], words[j] = words[j], words[i]
}
return strings.Join(words, " ")
}
func main() {
fmt.Println(reverse_words("one two three"))
}
Another hack is to use built-in language features, for example, defer:
package main
import "fmt"
func main() {
var name string
fmt.Scanln(&name)
for _, char := range []rune(name) {
defer fmt.Printf("%c", char) // <-- LIFO does it all for you
}
}
For simple strings it possible to use such construction:
func Reverse(str string) string {
if str != "" {
return Reverse(str[1:]) + str[:1]
}
return ""
}
For Unicode strings it might look like this:
func RecursiveReverse(str string) string {
if str == "" {
return ""
}
runes := []rune(str)
return RecursiveReverse(string(runes[1:])) + string(runes[0])
}
A version which I think works on unicode. It is built on the utf8.Rune functions:
func Reverse(s string) string {
b := make([]byte, len(s));
for i, j := len(s)-1, 0; i >= 0; i-- {
if utf8.RuneStart(s[i]) {
rune, size := utf8.DecodeRuneInString(s[i:len(s)]);
utf8.EncodeRune(rune, b[j:j+size]);
j += size;
}
}
return string(b);
}
rune is a type, so use it. Moreover, Go doesn't use semicolons.
func reverse(s string) string {
l := len(s)
m := make([]rune, l)
for _, c := range s {
l--
m[l] = c
}
return string(m)
}
func main() {
str := "the quick brown 狐 jumped over the lazy 犬"
fmt.Printf("reverse(%s): [%s]\n", str, reverse(str))
}
try below code:
package main
import "fmt"
func reverse(s string) string {
chars := []rune(s)
for i, j := 0, len(chars)-1; i < j; i, j = i+1, j-1 {
chars[i], chars[j] = chars[j], chars[i]
}
return string(chars)
}
func main() {
fmt.Printf("%v\n", reverse("abcdefg"))
}
for more info check http://golangcookbook.com/chapters/strings/reverse/
and http://www.dotnetperls.com/reverse-string-go
func reverseString(someString string) string {
runeString := []rune(someString)
var reverseString string
for i := len(runeString)-1; i >= 0; i -- {
reverseString += string(runeString[i])
}
return reverseString
}
Strings are immutable object in golang, unlike C inplace reverse is not possible with golang.
With C , you can do something like,
void reverseString(char *str) {
int length = strlen(str)
for(int i = 0, j = length-1; i < length/2; i++, j--)
{
char tmp = str[i];
str[i] = str[j];
str[j] = tmp;
}
}
But with golang, following one, uses byte to convert the input into bytes first and then reverses the byte array once it is reversed, convert back to string before returning. works only with non unicode type string.
package main
import "fmt"
func main() {
s := "test123 4"
fmt.Println(reverseString(s))
}
func reverseString(s string) string {
a := []byte(s)
for i, j := 0, len(s)-1; i < j; i++ {
a[i], a[j] = a[j], a[i]
j--
}
return string(a)
}
Here is yet another solution:
func ReverseStr(s string) string {
chars := []rune(s)
rev := make([]rune, 0, len(chars))
for i := len(chars) - 1; i >= 0; i-- {
rev = append(rev, chars[i])
}
return string(rev)
}
However, yazu's solution above is more elegant since he reverses the []rune slice in place.

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