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
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);
其他回答
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];
}
其中一种方法也可以是在使用分割法后再使用还原法进行逆向。
函数reverse(str) { 返回str.split(”)。reduce((rev, currentChar) => currentChar + rev, "); } console.log(反向('苹果')); console.log(反向('你好')); console.log(反向(“问候!”);
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'
看来我已经迟到3年了…
不幸的是,正如已经指出的那样,你不能。参见JavaScript字符串是不可变的吗?我需要一个“字符串生成器”在JavaScript?
你能做的下一个最好的事情是创建一个“视图”或“包装器”,它接受一个字符串并重新实现你正在使用的字符串API的任何部分,但假装字符串是反向的。例如:
var identity = function(x){return x};
function LazyString(s) {
this.original = s;
this.length = s.length;
this.start = 0; this.stop = this.length; this.dir = 1; // "virtual" slicing
// (dir=-1 if reversed)
this._caseTransform = identity;
}
// syntactic sugar to create new object:
function S(s) {
return new LazyString(s);
}
//We now implement a `"...".reversed` which toggles a flag which will change our math:
(function(){ // begin anonymous scope
var x = LazyString.prototype;
// Addition to the String API
x.reversed = function() {
var s = new LazyString(this.original);
s.start = this.stop - this.dir;
s.stop = this.start - this.dir;
s.dir = -1*this.dir;
s.length = this.length;
s._caseTransform = this._caseTransform;
return s;
}
//We also override string coercion for some extra versatility (not really necessary):
// OVERRIDE STRING COERCION
// - for string concatenation e.g. "abc"+reversed("abc")
x.toString = function() {
if (typeof this._realized == 'undefined') { // cached, to avoid recalculation
this._realized = this.dir==1 ?
this.original.slice(this.start,this.stop) :
this.original.slice(this.stop+1,this.start+1).split("").reverse().join("");
this._realized = this._caseTransform.call(this._realized, this._realized);
}
return this._realized;
}
//Now we reimplement the String API by doing some math:
// String API:
// Do some math to figure out which character we really want
x.charAt = function(i) {
return this.slice(i, i+1).toString();
}
x.charCodeAt = function(i) {
return this.slice(i, i+1).toString().charCodeAt(0);
}
// Slicing functions:
x.slice = function(start,stop) {
// lazy chaining version of https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Global_Objects/Array/slice
if (stop===undefined)
stop = this.length;
var relativeStart = start<0 ? this.length+start : start;
var relativeStop = stop<0 ? this.length+stop : stop;
if (relativeStart >= this.length)
relativeStart = this.length;
if (relativeStart < 0)
relativeStart = 0;
if (relativeStop > this.length)
relativeStop = this.length;
if (relativeStop < 0)
relativeStop = 0;
if (relativeStop < relativeStart)
relativeStop = relativeStart;
var s = new LazyString(this.original);
s.length = relativeStop - relativeStart;
s.start = this.start + this.dir*relativeStart;
s.stop = s.start + this.dir*s.length;
s.dir = this.dir;
//console.log([this.start,this.stop,this.dir,this.length], [s.start,s.stop,s.dir,s.length])
s._caseTransform = this._caseTransform;
return s;
}
x.substring = function() {
// ...
}
x.substr = function() {
// ...
}
//Miscellaneous functions:
// Iterative search
x.indexOf = function(value) {
for(var i=0; i<this.length; i++)
if (value==this.charAt(i))
return i;
return -1;
}
x.lastIndexOf = function() {
for(var i=this.length-1; i>=0; i--)
if (value==this.charAt(i))
return i;
return -1;
}
// The following functions are too complicated to reimplement easily.
// Instead just realize the slice and do it the usual non-in-place way.
x.match = function() {
var s = this.toString();
return s.apply(s, arguments);
}
x.replace = function() {
var s = this.toString();
return s.apply(s, arguments);
}
x.search = function() {
var s = this.toString();
return s.apply(s, arguments);
}
x.split = function() {
var s = this.toString();
return s.apply(s, arguments);
}
// Case transforms:
x.toLowerCase = function() {
var s = new LazyString(this.original);
s._caseTransform = ''.toLowerCase;
s.start=this.start; s.stop=this.stop; s.dir=this.dir; s.length=this.length;
return s;
}
x.toUpperCase = function() {
var s = new LazyString(this.original);
s._caseTransform = ''.toUpperCase;
s.start=this.start; s.stop=this.stop; s.dir=this.dir; s.length=this.length;
return s;
}
})() // end anonymous scope
演示:
> r = S('abcABC')
LazyString
original: "abcABC"
__proto__: LazyString
> r.charAt(1); // doesn't reverse string!!! (good if very long)
"B"
> r.toLowerCase() // must reverse string, so does so
"cbacba"
> r.toUpperCase() // string already reversed: no extra work
"CBACBA"
> r + '-demo-' + r // natural coercion, string already reversed: no extra work
"CBAcba-demo-CBAcba"
最重要的是——下面是用纯数学来完成的,每个角色只访问一次,而且是在必要的时候:
> 'demo: ' + S('0123456789abcdef').slice(3).reversed().slice(1,-1).toUpperCase()
"demo: EDCBA987654"
> S('0123456789ABCDEF').slice(3).reversed().slice(1,-1).toLowerCase().charAt(3)
"b"
如果应用于一个非常大的字符串,如果您只取其中相对较小的部分,则可以节省大量的时间。
Whether this is worth it (over reversing-as-a-copy like in most programming languages) highly depends on your use case and how efficiently you reimplement the string API. For example if all you want is to do string index manipulation, or take small slices or substrs, this will save you space and time. If you're planning on printing large reversed slices or substrings however, the savings may be small indeed, even worse than having done a full copy. Your "reversed" string will also not have the type string, though you might be able to fake this with prototyping.
The above demo implementation creates a new object of type ReversedString. It is prototyped, and therefore fairly efficient, with almost minimal work and minimal space overhead (prototype definitions are shared). It is a lazy implementation involving deferred slicing. Whenever you perform a function like .slice or .reversed, it will perform index mathematics. Finally when you extract data (by implicitly calling .toString() or .charCodeAt(...) or something), it will apply those in a "smart" manner, touching the least data possible.
注意:上面的字符串API是一个例子,可能不能完美地实现。你也可以只使用你需要的1-2个函数。
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
推荐文章
- 给一个数字加上st, nd, rd和th(序数)后缀
- 如何以编程方式触发引导模式?
- 如何分割逗号分隔的字符串?
- Java字符串—查看字符串是否只包含数字而不包含字母
- setTimeout带引号和不带括号的区别
- 在JS的Chrome CPU配置文件中,'self'和'total'之间的差异
- 用javascript检查输入字符串中是否包含数字
- 如何使用JavaScript分割逗号分隔字符串?
- 在Javascript中~~(“双波浪号”)做什么?
- 谷歌chrome扩展::console.log()从后台页面?
- 未捕获的SyntaxError:
- [].slice的解释。调用javascript?
- jQuery日期/时间选择器
- 我如何预填充一个jQuery Datepicker文本框与今天的日期?
- 数组的indexOf函数和findIndex函数的区别