详解Go语言如何实现字符串切片反转函数

Python 中的 reverse 函数

reverse()

>>> myList = [2022, 2021, 2008, 2012]
>>> myList.reverse()
>>> print("Reversed List:", myList)
Reversed List: [2012, 2008, 2021, 2022]
>>> 

实现一个 reverse 反转函数

reversesort

利用两个切片实现

设计思想：

• 确定切片长度
• 获取最后一个元素
• 以相反的顺序在新切片中添加最后一个元素到第一个位置

package main

import "fmt"

func main() {
    s := []string{"hello", "foo", "bar", "go", "abc", "zzz"}

    // 定义新的反转切片
    reverseOfS := []string{}

    // 遍历原切片 s
    for i := range s {
        reverseOfS = append(reverseOfS, s[len(s)-1-i])
    }
    fmt.Println(reverseOfS)
}

运行结果：

[zzz abc go bar foo hello]

O(n)

前后两两原地交换

reverse

func reverse(s []string) []string {
    for i := 0; i < len(s)/2; i++ {
        j := len(s) - i - 1
        s[i], s[j] = s[j], s[i]
    }
    return  s
}

这个函数可以通过更简短的实现，通过 Go 内部的操作进行循环：

package main

import "fmt"

func reverse(s []string) []string {
    for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
        s[i], s[j] = s[j], s[i]
    }
    return s
}

func main() {
    s := []string{"hello", "foo", "bar", "go", "abc", "zzz"}

    reverseOfS := reverse(s)

    fmt.Println(reverseOfS)
}

执行结果：

[zzz abc go bar foo hello]

reverse[]string

package main

import "fmt"

func reverse(s []string) {
    for i := 0; i < len(s)/2; i++ {
        j := len(s) - i - 1
        s[i], s[j] = s[j], s[i]
    }
}

func main() {
    s := []string{"hello", "foo", "bar", "go", "abc", "zzz"}

    reverse(s)
    fmt.Printf("%v\n", s)
}

reverse()reverse()

反转为原切片的副本

reverse

package main

import "fmt"

func reverse(s []string) []string {

    newS := make([]string, len(s))
    for i, j := 0, len(s)-1; i <= j; i, j = i+1, j-1 {
        newS[i], newS[j] = s[j], s[i]
    }
    return newS
}

func main() {
    s := []string{"hello", "foo", "bar", "go", "abc", "zzz"}

    fmt.Printf("原字符串切片：%v\n", s)
    fmt.Printf("反转后的切片：%v\n", reverse(s))
}

运行结果：

原字符串切片：[hello foo bar go abc zzz]
反转后的切片：[zzz abc go bar foo hello]

可以看到，原切片是没有变化的。

append

func reverse(s []string) []string {

    newS := make([]string, 0, len(s))

    for i := len(s)-1; i >= 0; i-- {
        newS = append(newS, s[i])
    }
    return newS
}

运行结果图如下：

总结

reverse()

reverse