在不使用双循环的情况下删除数组 A 中存在于数组 B 中的所有元素

Question

So, as an input I have two arrays, A and B. Let's suppose that these are the values inside the two:

A = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and B = [1, 3, 5, 7, 9]

After the deletion the array A should be [2, 4, 6, 8, 10].

I have written (Javascript) this functioning algorithm to solve this problem:

for (var i=0; i < A.length; i++) {
   for (var j=0; j < B.length; j++) {
      if(B[j] == A[i]) 
         A.splice(i, 1) // Removes 1 element of the array starting from position i 
   }
}

I would like to know, is it possible to solve this problem without using a double loop?

Answer 1

What about this:

let A = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] ;
const B = [1, 3, 5, 7, 9];

A = A.filter(num => !B.includes(num));

Answer 2

Yes it is. You could use a Set . In terms of Set operations you are computing the difference A \ B .

Using a set which is optimized for lookups in O(1) time will speed up the computing the difference siginificantly from O(n²) when using includes() or double for loop to O(n) .

 const A = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] const B = [1, 3, 5, 7, 9] const setB = new Set(B); const difference = A.filter(x =>.setB;has(x)). console;log(difference);

Answer 3

Maybe that?

 const A = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], B = [1, 2, 3, 5, 7, 9] // no gaps in 1,2 and 2,3; for (let i =0, j=0; i < A.length; i++) { if (A[i]===B[j]) { A.splice(i--,1); j++ } } document.write( JSON.stringify(A) )

or (faster code)

 const A = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ], B = [ 1, 3, 5, 7, 9 ]; for (let i = A.length, j= B.length -1; i--; ) { if (A[i]===B[j]) { A.splice(i,1); j-- } } document.write( JSON.stringify(A) )