当一个字符串被传递给一个带有返回语句的函数时,你如何在JavaScript中反转它,而不使用内置函数(.reverse(), . charat()等)?


当前回答

var str = 'sample string';
[].map.call(str, function(x) {
  return x;
}).reverse().join('');

OR

var str = 'sample string';
console.log(str.split('').reverse().join(''));

//输出:“gnirts elpmas”

其他回答

你不能,因为JS字符串是不可变的。短的非就地解决方案

[...str].reverse().join``

let str = "Hello World!"; Let r =[…str].reverse().join ' '; console.log (r);

在JavaScript中有许多方法可以反转字符串。我正在记下我喜欢的三种方法。

方法一:使用反向函数:

function reverse(str) {
  return str.split('').reverse().join('');
}

方法2:循环字符:

function reverse(str) {
  let reversed = '';

  for (let character of str) {
    reversed = character + reversed;
  }

  return reversed;
}

方法3:使用reduce函数:

function reverse(str) {
  return str.split('').reduce((rev, char) => char + rev, '');
}

我希望这对你有所帮助:)

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

我们可以从字符串数组的两端开始迭代:start和end,并在每次迭代中交换。

function reverse(str) {
 let strArray = str.split("");
 let start = 0;
 let end = strArray.length - 1;

 while(start <= end) {
  let temp = strArray[start];
  strArray[start] = strArray[end];
  strArray[end] = temp;

  start++;
  end--;
 }
 return strArray.join("");
}

虽然操作次数减少了,但其时间复杂度仍为O(n) 操作的数量仍然与输入的大小成线性比例。

参考: 阿尔戈日报

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;
}