javascript算法-返回第一个数组中第二个不存在的元素

Question

Hi I am stuck on this algorithm.

Can anyone see why this is not working for all tests:

Instructions: Your goal in this kata is to implement a difference function, which subtracts one list from another and returns the result. It should remove all values from list a, which are present in list b.

my attempt:

function arrayDiff(a, b) {
  for(let i = 0; i < a.length; i++){
    for(let j = 0; j < b.length; j++){
      if(a[i] === b[j]){
        a.splice(i,1 );
      
      } 
    } 
  }
   return a;
}

When i pass this to my function it works:

arrayDiff([1,8,2], []) it passes and returns []

When i pass this it fails:

arrayDiff([1,2,2], [2]) it fails and returns [1, 2] it should only return [1]

UPDATE * THIS IS THE QUESTION, not sure if that helps?

Your goal in this kata is to implement a difference function, which subtracts one list from another and returns the result.

It should remove all values from list a, which are present in list b.

arrayDiff([1,2],[1]) == [2] If a value is present in b, all of its occurrences must be removed from the other:

arrayDiff([1,2,2,2,3],[2]) == [1,3]

Answer 1

You need to decrement i after splicing the array, since the array got one shorter.

function arrayDiff(a, b) {
  for (let i = 0; i < a.length; i++) {
    for (let j = 0; j < b.length; j++) {
      if (a[i] === b[j]) {
        a.splice(i, 1);
        i--;
      }
    }
  }
  return a;
}

Edit: that said, a much more elegant solution to this problem would be to solve it functionally.

function arrayDiff(a, b) {
    return a.filter(elem => !b.includes(elem))
}

Answer 2

You can try using Array.Filter method.

    function arrayDiff(a, b) {
      return a.filter(value => !b.includes(value));
    }

Answer 3

with one for loop

function arrayDiff(a, b) {
  for (let i = 0; i < a.length; i++) {
    if(b.indexOf(a[i]) !== -1) {
       a.splice(i, 1);
       i--
    
     }
 }
   return a
}

simple way

function arrayDiff(a, b) {
   return a.filter(item => b.indexOf(item) === -1);
}