我随机来到这里，从来没有理由在javascript中重复字符。

我对artistoex的做法印象深刻，对结果感到失望。我注意到最后一个串连接是不必要的，正如丹尼斯也指出的那样。

我注意到一些更多的东西，当玩抽样disfate放在一起。

The results varied a fair amount often favoring the last run and similar algorithms would often jockey for position. One of the things I changed was instead of using the JSLitmus generated count as the seed for the calls; as count was generated different for the various methods, I put in an index. This made the thing much more reliable. I then looked at ensuring that varying sized strings were passed to the functions. This prevented some of the variations I saw, where some algorithms did better at the single chars or smaller strings. However the top 3 methods all did well regardless of the string size.

分叉测试集

http://jsfiddle.net/schmide/fCqp3/134/

// repeated string
var string = '0123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789';
// count paremeter is changed on every test iteration, limit it's maximum value here
var maxCount = 200;

var n = 0;
$.each(tests, function (name) {
    var fn = tests[name];
    JSLitmus.test(++n + '. ' + name, function (count) {
        var index = 0;
        while (count--) {
            fn.call(string.slice(0, index % string.length), index % maxCount);
            index++;
        }
    });
    if (fn.call('>', 10).length !== 10) $('body').prepend('<h1>Error in "' + name + '"</h1>');
});

JSLitmus.runAll();

然后我加入了丹尼斯的解决方案，并决定看看我是否能找到更多的方法。

由于javascript不能真正优化，提高性能的最好方法是手动避免一些东西。如果我把前4个琐碎的结果从循环中取出，我可以避免2-4个字符串存储，并将最后的存储直接写入结果。

// final: growing pattern + prototypejs check (count < 1)
'final avoid': function (count) {
    if (!count) return '';
    if (count == 1) return this.valueOf();
    var pattern = this.valueOf();
    if (count == 2) return pattern + pattern;
    if (count == 3) return pattern + pattern + pattern;
    var result;
    if (count & 1) result = pattern;
    else result = '';
    count >>= 1;
    do {
        pattern += pattern;
        if (count & 1) result += pattern;
        count >>= 1;
    } while (count > 1);
    return result + pattern + pattern;
}

这比丹尼斯的修复方案平均提高了1-2%。然而，不同的运行和不同的浏览器会显示相当大的差异，这额外的代码可能不值得在前面的两种算法上付出努力。

一个图表

编辑:我这样做主要是在chrome浏览器。Firefox和IE通常对Dennis的偏爱程度为百分之几。

要在指定次数内重复一个字符串，可以使用JavaScript中内置的repeat()方法。

下面是一个重复以下字符串4次的例子:

const name = "king";

const repeat = name.repeat(4);

console.log(repeat);

输出:

"kingkingkingking"

或者我们可以创建自己版本的repeat()函数，如下所示:

function repeat(str, n) {
  if (!str || !n) {
    return;
  }

 let final = "";
  while (n) {
    final += s;
    n--;
  }
  return final;
}

console.log(repeat("king", 3))

(最初发布于https://reactgo.com/javascript-repeat-string/)

在ES8中，你也可以使用padStart或padEnd。如。

var str = 'cat';
var num = 23;
var size = str.length * num;
"".padStart(size, str) // outputs: 'catcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcatcat'

ES-Next有很多方法

1. ES2015/ES6已经实现了这个repeat()方法!

/** * str: String * count: Number */ const str = `hello repeat!\n`, count = 3; let resultString = str.repeat(count); console.log(`resultString = \n${resultString}`); /* resultString = hello repeat! hello repeat! hello repeat! */ ({ toString: () => 'abc', repeat: String.prototype.repeat }).repeat(2); // 'abcabc' (repeat() is a generic method) // Examples 'abc'.repeat(0); // '' 'abc'.repeat(1); // 'abc' 'abc'.repeat(2); // 'abcabc' 'abc'.repeat(3.5); // 'abcabcabc' (count will be converted to integer) // 'abc'.repeat(1/0); // RangeError // 'abc'.repeat(-1); // RangeError

2. ES2017/ES8新增String.prototype.padStart()

Const STR = 'abc '; Const times = 3; const newStr = str. padstartlength * times, str.toUpperCase()); console.log(' newStr = '， newStr); // "newStr =" "ABC ABC ABC "

3.ES2017/ES8新增String.prototype.padEnd()

Const STR = 'abc '; Const times = 3; const newStr = str. padend (str。length * times, str.toUpperCase()); console.log(' newStr = '， newStr); // "newStr =" "abc abc abc "

refs

http://www.ecma-international.org/ecma-262/6.0/#sec-string.prototype.repeat

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/repeat

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padEnd

function repeat(s, n) { var r=""; for (var a=0;a<n;a++) r+=s; return r;}

新读者注意:这个答案很老，而且不太实用——它只是“聪明”，因为它使用Array的东西来获取 串事情做完了。当我写“更少的过程”时，我肯定是指 “更少的代码”，因为，正如其他人在随后的回答中指出的那样，它 表演像猪一样。所以如果你看重速度就不要用它。

我将把这个函数直接放到String对象上。与其创建一个数组，填充它，然后用一个空字符连接它，不如创建一个适当长度的数组，然后用你想要的字符串连接它。同样的结果，更少的过程!

String.prototype.repeat = function( num )
{
    return new Array( num + 1 ).join( this );
}

alert( "string to repeat\n".repeat( 4 ) );