Reverse numbers in function without using ".reverse" method in Javascript

Question

function revertNumbers(...numberArray) {
   let rev = [];
   for(let i = 0; i <numberArray.length; i++)
   {
     rev.push(numberArray[i])
   }
   return rev.reverse();
}

console.log("revertNumbers", revertNumbers(0, 1, 2, 3, 4, 5, 6, 7, 8, 9) === "9,8,7,6,5,4,3,2,1,0");

Can you please show me the how to reverse number in this code that the statement will be true? Also without using .reverse method. Is it possible to make it in another for loop by just changing this statement:

(let i = 0; i <numberArray.length; i++)

Technically you can keep the loop, just use rev.unshift(numberArray[i]). However unshift is slower than push. — tevemadar
– tevemadar, Commented Dec 31, 2019 at 14:26

Maheer Ali · Accepted Answer · 2019-12-31 14:19:40Z

3

You just need to reverse the direction of your loop. Means start i with last index and then gradually decrease it to 0

function revertNumbers(...numberArray) {
   let rev = [];
   for(let i = numberArray.length - 1; i >= 0; i--)
   {
     rev.push(numberArray[i])
   }
   return rev.join(",")
}

console.log("revertNumbers", revertNumbers(0, 1, 2, 3, 4, 5, 6, 7, 8, 9) === "9,8,7,6,5,4,3,2,1,0");

This is also a good use case of reduceRight()

const revertNumbers = (...arr) => arr.reduceRight((ac, a) => ([...ac, a]), []).join(',')

console.log("revertNumbers", revertNumbers(0, 1, 2, 3, 4, 5, 6, 7, 8, 9) === "9,8,7,6,5,4,3,2,1,0");

answered Dec 31, 2019 at 14:19

Maheer Ali

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1 Comment

Stephen M Irving Over a year ago

It is certainly clever, but also a bit hacky to use reduceRight for this purpose, no? The actual purpose of the code is obscured by the implementation method chosen. The first method was the best and I upvoted your answer for it. You could also write the for loop in the first example in a shorter way, like this: for (let i = numberArray.length; i--;) { ... }

Stephen M Irving · Accepted Answer · 2020-01-03 22:02:43Z

Here is a solution that manipulates the array in place, and only has to traverse half of the original array in order to reverse it. This ekes out some modest performance gains compared to other answers in this thread (my function narrowly beats out or matches the speed of even the native reverse method in ops/sec), but micro-optimizations are largely irrelevant for this problem unless you are talking about a truly massive list of numbers.

Nonetheless, here is my answer:

const revNums = (...numArray) => {
  for (
    let arrLen = numArray.length,
        breakPoint = ((arrLen / 2)|0) - 1,
        i = arrLen,
        k = 0,
        temp;
    --i !== breakPoint;
    ++k
  ) {
    temp = numArray[i];
    numArray[i] = numArray[k];
    numArray[k] = temp;
  }

  return numArray.join(',');
};

Mr. Polywhirl · Accepted Answer · 2019-12-31 14:24:40Z

0

You could reduce the original array and unshift each element onto the new array.

function revertNumbers(...numberArray) {
  return numberArray.reduce((r, e) => { r.unshift(e); return r }, []).join(',')
}

console.log("revertNumbers", revertNumbers(0, 1, 2, 3, 4, 5, 6, 7, 8, 9) === "9,8,7,6,5,4,3,2,1,0")

If you don't want to unshift, you can concat in reverse.

function revertNumbers(...numberArray) {
  return numberArray.reduce((r, e, i, a) => r.concat(a[a.length - i - 1]), []).join(',')
}

console.log("revertNumbers", revertNumbers(0, 1, 2, 3, 4, 5, 6, 7, 8, 9) === "9,8,7,6,5,4,3,2,1,0")

answered Dec 31, 2019 at 14:24

Mr. Polywhirl

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Comments

Nina Scholz · Accepted Answer · 2019-12-31 14:40:59Z

0

You could take a value and the rest of the arguments and use a recursive approach to get a reversed array of arguments.

function revertNumbers(v, ...rest) {
    return rest.length
        ? [...revertNumbers(...rest), v]
        : [v];
}

console.log(revertNumbers(0, 1, 2, 3, 4, 5, 6, 7, 8, 9));

answered Dec 31, 2019 at 14:40

Nina Scholz

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Comments

Mister Jojo · Accepted Answer · 2019-12-31 17:33:20Z

It is to ask how many solutions can be possible ?
here are 3 of them...

"use strict";

const targetString = '9,8,7,6,5,4,3,2,1,0'
  ;
function soluce_1(...numberArray)
  {
  const rev = [];
  for( let i=numberArray.length;i--;) { rev.push(numberArray[i])  }
  return rev.join(',')
  }
function soluce_2(...numberArray)
  {
  const rev = [];
  let   pos = numberArray.length;
  for(let N in numberArray) { rev[--pos] = N }
  return rev.join(",")
  }
function soluce_3(...numberArray)
  {
  const rev = [];
  while(numberArray.length) { rev.push(numberArray.pop()) }
  return rev.join(',')
  }


console.log('soluce_1 ->', (soluce_1(0,1,2,3,4,5,6,7,8,9)===targetString) );
console.log('soluce_2 ->', (soluce_2(0,1,2,3,4,5,6,7,8,9)===targetString) );
console.log('soluce_3 ->', (soluce_3(0,1,2,3,4,5,6,7,8,9)===targetString) );

And yes, I code in Whitesmiths style, please respect this (the reason for the downVote for correct answers ?)

https://en.wikipedia.org/wiki/Indentation_style#Whitesmiths_style

Kosh · Accepted Answer · 2020-01-03 23:41:53Z

0

You might reverse like this:

function reverse(...a) {
  const h = a.length >> 1, l = a.length-1;
  for (let i = 0; i < h; ++i) [a[i], a[l-i]] = [a[l-i], a[i]];
  return a;
}

console.log(reverse(0, 1, 2, 3, 4, 5, 6, 7, 8, 9).join(','));

answered Jan 3, 2020 at 23:41

Kosh

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