
Solution to Day 5 of 7 Days of JS
3 December 2019
As before, I first made it with a while loop:
function positionOf(n,array){
let i = 0;
let count = array.length;
while (i < count){
if (array[i] == n){
return i;
} else {
i++;
}
}
return -1;
}
console.log(positionOf(5,[-5,0,50,5,35]))
And then, a for loop:
function positionOf1(n,array){
let count = array.length;
for (let i = 0; i < count; i++){
if (array[i] == n){
return i;
}
}
return -1;
}
console.log(positionOf1(3,[-3,0,3,5,35]))
Do you have a better solution? Share it in the comments!
See you on day 6!