CODEWITHSUNDEEP
javascriptnode.jsarrays
BaRzz007
BaRzz007

Reputation: 3

How do i reverse an array in javaScript without .reverse()

I'm trying to reverse the contents of an array. My approach works well when the contents of the said array are of same type. but once they're of different types it just doesn't work.
Constraint is that i can't use the .reverse() method and if possible, without creating a new array.
The answer in this question is close to what i want but i don't understand it.

Here is my code...

reverse.js

#!/usr/bin/node
exports.rev = function (list) {
  for (let i = 0, j = (list.length - 1); i <= (list.length / 2); i++, j--) {
    [list[i], list[j]] = [list[j], list[i]];
  }
  return (list);
};

main.js

#!/usr/bin/node
const rev = require('./reverse').rev;

console.log(rev([1, 2, 3, 4, 5]));
console.log(rev(["School", 89, { id: 12 }, "String"]));

Expected:
[5, 4, 3, 2, 1]
["String", { id: 12 }, 89, "School"]

What I got:
[5, 4, 3, 2, 1]
["String", 89, { id: 12 }, "School"]

Upvotes: -1

Views: 112

Answers (5)

malarres
malarres

Reputation: 2946

Already good answers (and examples of arrays) here

My try with native methods:

const rev = (arr) => {
  len = arr.length;
  for (i = 1; i<= len; i++) arr.splice(len-i,0,arr.shift());
}

const array = ["School", 89, { id: 12 }, "String"];

console.log("original",array);
rev(array);
console.log("reversed",array);

Upvotes: 0

Rajesh
Rajesh

Reputation: 24915

You can user a recursion to reverse

You can use Spread operator(...) along with distructure assignment for that.

function rev(arr) {
  const [first, ...rest] = arr
  return arr.length > 1 ? [...rev(rest), first] : arr
}

console.log(rev([1, 2, 3, 4, 5]));
console.log(rev(["School", 89, { id: 12 }, "String"]));

const [first, ...rest] = arr is a shortcut for:

const first = arr[0]
const rest = arr.slice(1)

However, ...rev(rest) will spread items of returned array and spread them. So, [...rev(rest), first] will keep the output of rev first and then push first at the end of array.

If you are comfortable with mutation of original array, try this

function rev(arr) {
  return arr.length > 1 ? [arr.pop(), ...rev(arr)] : arr
}

console.log(rev([1, 2, 3, 4, 5]));
console.log(rev(["School", 89, { id: 12 }, "String"]));

Upvotes: 0

davidhuerta
davidhuerta

Reputation: 44

You can try with a for loop like this:

let arr = ["School", 89, { id: 12 }, "String"];

let newArr = [];

for (let i = arr.length - 1; i >= 0; i--) {
  newArr.push(arr[i])
}
console.log(newArr);

Expected output: ["String", [object Object] { id: 12 }, 89, "School"]

Upvotes: 0

Nick Vu
Nick Vu

Reputation: 15520

I've found out that your code almost works. You just need to modify the condition a bit to

i < (list.length / 2) //not `<=`


function rev(list) {
  for (let i = 0, j = (list.length - 1); i < (list.length / 2); i++, j--) {
    [list[i], list[j]] = [list[j], list[i]];
  }
  return (list);
};

console.log(rev([1, 2, 3, 4, 5]));
console.log(rev(["School", 89, { id: 12 }, "String"]));

Upvotes: 1

David Ho
David Ho

Reputation: 287

let array = [1, 2, 3, 4, 5]
let reverse = [];

for (var i = array.length - 1; i >= 0; i--){
    reverse.push(array[i]);
}

Upvotes: 0

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