在ES6中，你还有一个选择

function reverseString (str) {
  return [...str].reverse().join('')
}

reverseString('Hello');

ES6

 function reverseString(str) {
     return [...str].reverse().join("");
 }

 console.log(reverseString("Hello")); // olleH

function reverseString(string) {
    var reversedString = "";
    var stringLength = string.length - 1;
    for (var i = stringLength; i >= 0; i--) {
        reversedString += string[i];
    }
    return reversedString;
}

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

//recursive implementation
function reverse(wrd) {
  const str =wrd[0]
  if(!wrd.length) {
    return wrd
  }
  return reverse(wrd.slice(1)) + str
}

使用内置函数反转字符串

函数reverse(str) { //使用split()方法返回一个新数组 //使用reverse()方法反转新创建的数组 //使用join()方法将数组中的所有元素连接成一个字符串 返回str.split (" ") .reverse () . join (" "); ｝ console.log(反向('你好'));

反转一个没有helper的字符串

函数reversedOf(str) { let newStr = "; For (let char of str) { newStr = char + newStr / /轮1:“h”+“= h,第二轮:“e”+“h”=“嗯”…等。 / / console.log (newStr); ｝ 返回newStr; ｝ console.log (reversedOf('你好'));