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
function reverse_string(string)
{
var string;
var len = string.length;
var stringExp = string.split('');
var i;
for (i = len-1; i >=0;i--)
{
var result = document.write(stringExp[i]);
}
return result;
}
reverse_string("This is a reversed string");
//outputs: gnirts描述一个sihT
其他回答
保持干燥和简单,傻!!
function reverse(s){
let str = s;
var reverse = '';
for (var i=str.length;i>0;i--){
var newstr = str.substring(0,i)
reverse += newstr.substr(-1,1)
}
return reverse;
}
在一次面试中,我被要求在不使用任何变量或本机方法的情况下反转一个字符串。这是我最喜欢的实现:
function reverseString(str) {
return str === '' ? '' : reverseString(str.slice(1)) + str[0];
}
使用Array函数,
String.prototype.reverse = function(){
return [].reduceRight.call(this, function(last, secLast){return last + secLast});
}
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
ES6
function reverseString(str) {
return [...str].reverse().join("");
}
console.log(reverseString("Hello")); // olleH