Fastest way to check if number is in array?

Question

I have to create a function that checks if number is in array. So I've tried this:

function getNumber(x, array) {
for (let i = 0; i < array.length; i++) {
    if (!x == array[i]) {
        console.log(false);
    } else if (x == array[i]) {
        console.log(true);
    }
}

getNumber(4, [5, 10, 2, 3, 5]);

It works just if x is in array but if it's not console doesn't show anything

I want to know if there is easier(faster) way to check it

Use `includes` for array – Rakesh Jakhar Mar 08 '20 at 13:44 — Rakesh Jakhar, Mar 08 '20 at 13:44

score 2 · Accepted Answer · answered Mar 08 '20 at 13:44

I think you can try with .includes() - that's definitely easier:

const array = [5, 10, 2, 3, 5];
const check1 = 4;
const check2 = 10;

const getNumber = (check, array) => {
  return array.includes(check);
}

console.log(getNumber(check1, array));
console.log(getNumber(check2, array));

I hope this helps!

score 0 · Answer 2 · answered Mar 08 '20 at 13:45

0

Use includes

var arr = [5, 10, 2, 3, 5];

if(arr.includes(21)){
 console.log(true);
}else{
  console.log(false);
}

answered Mar 08 '20 at 13:45

Rakesh Jakhar

6,298
2
9
20

score 0 · Answer 3 · answered Mar 08 '20 at 13:45

You could also use indexOf

    const array = [5, 10, 2, 3, 5];
    const check1 = 4;
    const check2 = 10;
    
    const getNumber = (check, array) => {
      return array.indexOf(check)>-1;
    }

console.log(getNumber(check1, array));
console.log(getNumber(check2, array));

Might be useful if you need higher browser coverage than includes https://caniuse.com/array-includes

Fastest way to check if number is in array?

3 Answers3