Move the zeros from an array at the end without using native functions-CodePudding

Having this array:

const myArry = [1, 543, 0, 232, 1, 45654, -5, 0, 7, 4, 0, 43, 77, 0, 77, 0]

It must be sorted in ascending order and place all 0s at the end, so the output should be:

[-5, 1, 1, 4, 7, 43, 77, 77, 232, 543, 45654, 0, 0, 0, 0, 0]

For sorting it's straightforward:

function sorting(arr) {

  for(let i = 0; i< arr.length; i  ) {
    for (let j = 0; j < arr.length - i -1; j  ) {
      if(arr[j 1] < arr[j]) {
        [arr[j 1], arr[j]]=[arr[j], arr[j 1]];
      }
    }
  }

  return arr;
}

But for moving those 0s I cannot find a solution to do it without native functions like push:

function moveZeros(arr) {
  let newArray = [];
  let counter = 0;
  for (let i = 0; i < arr.length; i  ) {
    if(arr[i] !== 0) {
      newArray.push(arr[i]);
    }
      else { counter  ; }
    }
  for (let j = 0; j < counter; j  ) {
    newArray.push(0);
  }
  return newArray;
}

Is there a way to do this? Also, if combining both methods into one

CodePudding user response：

When sorting numbers, to move all zeroes to the end, you have to give it more weight than any other number. Here, I "edge-cased" zero to always swap such that it moves to the end.

const myArry = [1, 543, 0, 232, 1, 45654, -5, 0, 7, 4, 0, 43, 77, 0, 77, 0]

function sorting(arr) {

  for(let i = 0; i< arr.length; i  ) {
    for (let j = 0; j < arr.length - i -1; j  ) {
      if(arr[j 1] < arr[j] && arr[j 1] != 0 || arr[j] == 0) {
        [arr[j 1], arr[j]]=[arr[j], arr[j 1]];
      }
    }
  }

  return arr;
}

console.log(''   sorting(myArry))

CodePudding user response：

You can do it with the sort function. It s much readable.

myArry.sort((a, b) => { 
  if(b === 0) return -Infinity; 
  if(a === 0) return Infinity; 
  return a -b 
})