我认为String.prototype.reverse是解决这个问题的好方法; 代码如下;

String.prototype.reverse = function() {
  return this.split('').reverse().join('');
}

var str = 'this is a good example for string reverse';
str.reverse();
-> "esrever gnirts rof elpmaxe doog a si siht";

你不能将字符串反向，但你可以使用这个:

String.prototype.reverse = function() { 返回this.split (" ") .reverse () . join (" "); ｝ var s = "ABCD"; S = S .reverse(); console.log(年代);

有多种方法，你可以检查下面的，

1. 传统的for循环(递增):

函数reverseString (str) { let stringRev =""; For(令i= 0;我< str.length;我+ +){ stringRev = str[i]+stringRev; ｝ 返回stringRev; ｝ alert (reverseString(“Hello World !”);

2. 传统的for循环(递减):

函数反向字符串（str）{ 让 revstr = “”; for（let i = str.length-1， i>=0， i--）{ Revstr = Revstr+ str[i]; } 返回修订版; } alert（reverseString（“Hello World！”））;

3.使用for-of循环

函数反向字符串（str）{ 让 strn =“”; for（let char of str）{ strn = char + strn; } 返回 STRN; } alert（reverseString（“Get Well soon”））;

4. 使用forEach/高阶数组方法:

函数反向字符串（str）{ 让 revSrring = “”; str.split（“”）.forEach（function（char）{ revSrring = char + revSrring; }); 返回转速Srring; } alert（reverseString（“Learning JavaScript”））;

5. ES6标准:

函数反向字符串（str）{ 让 revSrring = “”; str.split（“”）.forEach（char => revSrring = char + revSrring）; 返回转速Srring; } alert（reverseString（“Learning JavaScript”））;

6. 最新的方式:

函数反向字符串（str）{ return str.split（“”）.reduce（function（revString， char）{ 返回字符 + revString; }, ""); } alert（reverseString（“Learning JavaScript”））;

7. 你也可以用下面的方法得到结果，

函数反向字符串（str）{ return str.split（“”）.reduce（（revString， char）=> char + revString， “”）; } alert（reverseString（“Learning JavaScript”））;

String.prototype.reverse_string=function() {return this.split("").reverse().join("");}

or

String.prototype.reverse_string = function() {
    var s = "";
    var i = this.length;
    while (i>0) {
        s += this.substring(i-1,i);
        i--;
    }
    return s;
}

真正的答案是:你不能把它颠倒过来，但是你可以创建一个颠倒过来的新字符串。

Just as an exercise to play with recursion: sometimes when you go to an interview, the interviewer may ask you how to do this using recursion, and I think the "preferred answer" might be "I would rather not do this in recursion as it can easily cause a stack overflow" (because it is O(n) rather than O(log n). If it is O(log n), it is quite difficult to get a stack overflow -- 4 billion items could be handled by a stack level of 32, as 2 ** 32 is 4294967296. But if it is O(n), then it can easily get a stack overflow.

有时候面试官还是会问你，“作为练习，你为什么不用递归来写呢?”就是这样:

String.prototype.reverse = function() {
    if (this.length <= 1) return this;
    else return this.slice(1).reverse() + this.slice(0,1);
}

测试运行:

var s = "";
for(var i = 0; i < 1000; i++) {
    s += ("apple" + i);
}
console.log(s.reverse());

输出:

999elppa899elppa...2elppa1elppa0elppa

为了尝试获得堆栈溢出，我在谷歌Chrome中将1000更改为10000，它报告:

RangeError: Maximum call stack size exceeded

var str = "my name is saurabh ";
var empStr='',finalString='';
var chunk=[];
function reverse(str){
var i,j=0,n=str.length;
    for(i=0;i<n;++i){
        if(str[i]===' '){
            chunk[j]=empStr;
            empStr = '';
            j++;
        }else{
            empStr=empStr+str[i];
        }
    }
    for(var z=chunk.length-1;z>=0;z--){
        finalString = finalString +' '+ chunk[z];
        console.log(finalString);
    }
    return true;
}
reverse(str);