I'm studying JS and I have this exercise that is asking to reverse an array in place (without the use of a second array) and without the use of 'reverse'. Although I already have the solution to the exercise I don't understand why my solution does not work, here it is:
function reverseArrayInPlace (arr){
const k = arr[0];
while (arr[arr.length-1] !== k){
arr.unshift(arr.pop());
}
return arr;
}
console.log(reverseArrayInPlace(arr1));You take the end of the array and put it at the first position:
[1, 2, 3]
[3, 1, 2]
[2, 3, 1]
[1, 2, 3]
as you can see that actually doesnt reverse anything.
It will not work if your array contains duplicates of the first element. As you are taking the first element as key, whenever any duplicate element becomes the last element, your loop exits.
Try this, just check if the two elements being selected is equal or not, if equal do not swap else swap. Iterate till the pointer k is <= the pointer j.
function reverseArrayInPlace (arr){
let first = 0;
let last = arr.length - 1;
let k = first, j = last;
while(k <= j){
if(arr[k] !== arr[j]){
let temp = arr[k];
arr[k] = arr[j];
arr[j] = temp;
}
k++;
j--;
}
return arr;
}
arr1 = [1, 2, 3, 4];
console.log(reverseArrayInPlace(arr1));
arr1 = [1, 2, 3];
console.log(reverseArrayInPlace(arr1));This method will solve the problem without pop or unshift. Try this.
function reverseArray(array) {
for (let i = 0; i < Math.floor(array.length / 2); i++) {
let oldArray = array[i];
array[i] = array[array.length - 1 - i];
array[array.length - 1 - i] = oldArray;
}
return array;
}
console.log(reverseArray([1,2,3]));
[0, 1, 2, 0]? then the loop won't run asarr[0] === arr[arr.length - 1]...