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
//recursive implementation
function reverse(wrd) {
const str =wrd[0]
if(!wrd.length) {
return wrd
}
return reverse(wrd.slice(1)) + str
}
其他回答
下面是一个基本的ES6不可变的例子,没有使用Array.prototype.reverse:
//:: reverse = String ->字符串 const reverse = s => []. reduceright .使用实例调用(s, (a, b) => a + b) Console.log (reverse('foo')) // => 'oof' Console.log (reverse('bar')) // => 'rab' Console.log (reverse('foo-bar')) // => ' rabb -oof'
function reverseString(string) {
var reversedString = "";
var stringLength = string.length - 1;
for (var i = stringLength; i >= 0; i--) {
reversedString += string[i];
}
return reversedString;
}
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
这样的事情应该遵循最佳实践:
(function(){ 'use strict'; var str = "testing"; //using array methods var arr = new Array(); arr = str.split(""); arr.reverse(); console.log(arr); //using custom methods var reverseString = function(str){ if(str == null || str == undefined || str.length == 0 ){ return ""; } if(str.length == 1){ return str; } var rev = []; for(var i = 0; i < str.length; i++){ rev[i] = str[str.length - 1 - i]; } return rev; } console.log(reverseString(str)); })();
在ECMAScript 6中,你可以在不使用.split(") split方法的情况下更快地反转字符串,展开操作符如下所示:
var str = [...'racecar'].reverse().join('');