Javascript has a reverse() method that you can call in an array
var a = [3,5,7,8];
a.reverse(); // 8 7 5 3
Not sure if that's what you mean by 'libraries you can't use', I'm guessing something to do with practice. If that's the case, you can implement your own version of .reverse()
function reverseArr(input) {
var ret = new Array;
for(var i = input.length-1; i >= 0; i--) {
ret.push(input[i]);
}
return ret;
}
var a = [3,5,7,8]
var b = reverseArr(a);
Do note that the built-in .reverse() method operates on the original array, thus you don't need to reassign a.
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Javascript has a reverse() method that you can call in an array
var a = [3,5,7,8];
a.reverse(); // 8 7 5 3
Not sure if that's what you mean by 'libraries you can't use', I'm guessing something to do with practice. If that's the case, you can implement your own version of .reverse()
function reverseArr(input) {
var ret = new Array;
for(var i = input.length-1; i >= 0; i--) {
ret.push(input[i]);
}
return ret;
}
var a = [3,5,7,8]
var b = reverseArr(a);
Do note that the built-in .reverse() method operates on the original array, thus you don't need to reassign a.
Array.prototype.reverse()is all you need to do this work. See compatibility table.
var myArray = [20, 40, 80, 100];
var revMyArr = [].concat(myArray).reverse();
console.log(revMyArr);
// [100, 80, 40, 20]
Based on this setup:
var array = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
var length = array.length;
Array.reverse(); is the first or second slowest!
The benchmarks are here:
https://jsperf.com/js-array-reverse-vs-while-loop/9
Across browsers, swap loops are faster. There are two common types of swap algorithms (see Wikipedia), each with two variations.
The two types of swap algorithms are temporary swap and XOR swap.
The two variations handle index calculations differently. The first variation compares the current left index and the right index and then decrements the right index of the array. The second variation compares the current left index and the length divided by half and then recalculates the right index for each iteration.
You may or may not see huge differences between the two variations. For example, in Chrome 18, the first variations of the temporary swap and XOR swap are over 60% slower than the second variations, but in Opera 12, both variations of the temporary swap and XOR swap have similar performance.
Temporary swap:
First variation:
function temporarySwap(array)
{
var left = null;
var right = null;
var length = array.length;
for (left = 0, right = length - 1; left < right; left += 1, right -= 1)
{
var temporary = array[left];
array[left] = array[right];
array[right] = temporary;
}
return array;
}
Second variation:
function temporarySwapHalf(array)
{
var left = null;
var right = null;
var length = array.length;
for (left = 0; left < length / 2; left += 1)
{
right = length - 1 - left;
var temporary = array[left];
array[left] = array[right];
array[right] = temporary;
}
return array;
}
XOR swap:
First variation:
function xorSwap(array)
{
var i = null;
var r = null;
var length = array.length;
for (i = 0, r = length - 1; i < r; i += 1, r -= 1)
{
var left = array[i];
var right = array[r];
left ^= right;
right ^= left;
left ^= right;
array[i] = left;
array[r] = right;
}
return array;
}
Second variation:
function xorSwapHalf(array)
{
var i = null;
var r = null;
var length = array.length;
for (i = 0; i < length / 2; i += 1)
{
r = length - 1 - i;
var left = array[i];
var right = array[r];
left ^= right;
right ^= left;
left ^= right;
array[i] = left;
array[r] = right;
}
return array;
}
There is another swap method called destructuring assignment: http://wiki.ecmascript.org/doku.php?id=harmony:destructuring
Destructuring assignment:
First variation:
function destructuringSwap(array)
{
var left = null;
var right = null;
var length = array.length;
for (left = 0, right = length - 1; left < right; left += 1, right -= 1)
{
[array[left], array[right]] = [array[right], array[left]];
}
return array;
}
Second variation:
function destructuringSwapHalf(array)
{
var left = null;
var right = null;
var length = array.length;
for (left = 0; left < length / 2; left += 1)
{
right = length - 1 - left;
[array[left], array[right]] = [array[right], array[left]];
}
return array;
}
Right now, an algorithm using destructuring assignment is the slowest of them all. It is even slower than Array.reverse();. However, the algorithms using destructuring assignments and Array.reverse(); methods are the shortest examples, and they look the cleanest. I hope their performance gets better in the future.
Another mention is that modern browsers are improving their performance of array push and splice operations.
In Firefox 10, this for loop algorithm using array push and splice rivals the temporary swap and XOR swap loop algorithms.
for (length -= 2; length > -1; length -= 1)
{
array.push(array[length]);
array.splice(length, 1);
}
However, you should probably stick with the swap loop algorithms until many of the other browsers match or exceed their array push and splice performance.
In simple way you can do this using map.
let list = [10, 20, 30, 60, 90]
let reversedList = list.map((e, i, a)=> a[(a.length -1) -i]) // [90, 60...]
As long as you're dealing with simple ASCII characters, and you're happy to use built-in functions, this will work:
function reverse(s){
return s.split("").reverse().join("");
}
If you need a solution that supports UTF-16 or other multi-byte characters, be aware that this function will give invalid unicode strings, or valid strings that look funny. You might want to consider this answer instead.
The array expansion operator is Unicode aware:
function reverse(s){
return [...s].reverse().join("");
}
Another Unicode aware solution using split(), as explained on MDN, is to use a regexp with the u (Unicode) flag set as a separator.
function reverse(s){
return s.split(/(?:)/u).reverse().join("");
}
The following technique (or similar) is commonly used to reverse a string in JavaScript:
// Don’t use this!
var naiveReverse = function(string) {
return string.split('').reverse().join('');
}
In fact, all the answers posted so far are a variation of this pattern. However, there are some problems with this solution. For example:
naiveReverse('foo 𝌆 bar');
// → 'rab �� oof'
// Where did the `𝌆` symbol go? Whoops!
If you’re wondering why this happens, read up on JavaScript’s internal character encoding. (TL;DR: 𝌆 is an astral symbol, and JavaScript exposes it as two separate code units.)
But there’s more:
// To see which symbols are being used here, check:
// http://mothereff.in/js-escapes#1ma%C3%B1ana%20man%CC%83ana
naiveReverse('mañana mañana');
// → 'anãnam anañam'
// Wait, so now the tilde is applied to the `a` instead of the `n`? WAT.
A good string to test string reverse implementations is the following:
'foo 𝌆 bar mañana mañana'
Why? Because it contains an astral symbol (𝌆) (which are represented by surrogate pairs in JavaScript) and a combining mark (the ñ in the last mañana actually consists of two symbols: U+006E LATIN SMALL LETTER N and U+0303 COMBINING TILDE).
The order in which surrogate pairs appear cannot be reversed, else the astral symbol won’t show up anymore in the ‘reversed’ string. That’s why you saw those �� marks in the output for the previous example.
Combining marks always get applied to the previous symbol, so you have to treat both the main symbol (U+006E LATIN SMALL LETTER N) as the combining mark (U+0303 COMBINING TILDE) as a whole. Reversing their order will cause the combining mark to be paired with another symbol in the string. That’s why the example output had ã instead of ñ.
Hopefully, this explains why all the answers posted so far are wrong.
To answer your initial question — how to [properly] reverse a string in JavaScript —, I’ve written a small JavaScript library that is capable of Unicode-aware string reversal. It doesn’t have any of the issues I just mentioned. The library is called Esrever; its code is on GitHub, and it works in pretty much any JavaScript environment. It comes with a shell utility/binary, so you can easily reverse strings from your terminal if you want.
var input = 'foo 𝌆 bar mañana mañana';
esrever.reverse(input);
// → 'anañam anañam rab 𝌆 oof'
As for the “in-place” part, see the other answers.