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
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;
}
其他回答
你可以尝试这样做。我相信还有重构的空间。我无法使用分裂函数。也许有人知道不分裂的方法。
代码设置,可以把它放在你的。js库
使用它的代码(有客户端代码,只是因为它在浏览器中测试过):
var sentence = "My Stack is Overflowing."
document.write(sentence.reverseLetters() + '<br />');
document.write(sentence.reverseWords() + '<br />');
代码片段:
String.prototype.aggregate = function(vals, aggregateFunction) { var temp = ''; for (var i = vals.length - 1; i >= 0; i--) { temp = aggregateFunction(vals[i], temp); } return temp; } String.prototype.reverseLetters = function() { return this.aggregate(this.split(''), function(current, word) { return word + current; }) } String.prototype.reverseWords = function() { return this.aggregate(this.split(' '), function(current, word) { return word + ' ' + current; }) } var sentence = "My Stack is Overflowing." document.write(sentence.reverseLetters() + '<br />'); document.write(sentence.reverseWords() + '<br />');
字符串本身是不可变的,但是你可以用下面的代码轻松地创建一个反向副本:
function reverseString(str) {
var strArray = str.split("");
strArray.reverse();
var strReverse = strArray.join("");
return strReverse;
}
reverseString("hello");
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"
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;
}
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