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

我自己最初的尝试…

var str = "The Car";

function reverseStr(str) {
  var reversed = "";
  var len = str.length;
  for (var i = 1; i < (len + 1); i++) {  
    reversed += str[len - i];      
  }

  return reversed;
}

var strReverse = reverseStr(str);    
console.log(strReverse);
// "raC ehT"

http://jsbin.com/bujiwo/19/edit?js,console,output

其他回答

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;
}

好的,很简单,你可以创建一个简单的循环函数来为你反向字符串,而不需要使用reverse(), charAt()等,就像这样:

例如,你有这样一个字符串:

var name = "StackOverflow";

创建一个这样的函数,我称之为reverseString…

function reverseString(str) {
  if(!str.trim() || 'string' !== typeof str) {
    return;
  }
  let l=str.length, s='';
  while(l > 0) {
    l--;
    s+= str[l];
  }
  return s;
}

你可以这样称呼它:

reverseString(name);

结果是:

"wolfrevOkcatS"

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);

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
word.split('').reduce((acc, curr) => curr+""+acc)