从数组的末尾找到第 n 个元素

Question

i looked at this challenge from codesignal (nthelementfromtheend) and put my code (below) in a test site

function nthElementFromTheEnd(l, n) {
if (n > l.length){
    return -1;
}else{

// console.log();
let index = l.length - n;
// console.log(index);
// console.log(l[index]);
return l[index];
}
}

let l = [1, 2, 3, 4];
let n=7;
nthElementFromTheEnd(l, n);

results seem to pass the test site, but not codesignal.

open links below in new tab

challenge

tester

array length

Answer 1

You need to analyze the input that is coming into the function. l represents a singly-linked list. This doesn't exist natively in JavaScript, but it has been re-created using an object, as the comment describes:

// Singly-linked lists are already defined with this interface:
function ListNode(x) {
    this.value = x;
    this.next = null;
}

In the first test, the input that comes to the function looks like this:

ListNode {
    value: 1,
    next: ListNode {
        value: 2,
        next: ListNode {
            value: 3,
            next: null
        }
    }
}

So this is not as simple as returning a particular index from an array, because the function is not receiving an array but an object. You have to navigate the data structure continually checking for next values. There are probably more efficient ways to do this, but here's an example that at least passes the 8 sample tests:

function nthElementFromTheEnd(l, n) {
    let values = [];
    let node = l;

    while (node) {
        values.push(node.value);
        node = node.next;
    }

    let len = values.length;

    if (n > len) {
        return -1;
    } else {
        return values[len-n];
    }
}

Answer 2

The trick here is in the comment indicating the interface of a singly-linked lists.

// Singly-linked lists are already defined with this interface:
// function ListNode(x) {
//   this.value = x;
//   this.next = null;
// }
//

So you need to use l.next and l.value to navigate and get values from the linked list.

Here is a possible solution (not optimized):

function nthElementFromTheEnd(l, n) {
    // find the length of the linked list
    let len = 1;
    let c = l;
    while (c.next) {
        len++;
        c = c.next;
    }

    if (n > len) {
        return -1
    }
    else {
        // iterate linked list and get desired value (len-n)
        let i = 0;
        while (i < len-n){
            l = l.next;
            i++;
        }

        return l.value;
    }
}

Answer 3

function nthElementFromTheEnd(l, n) {
var input = l;
var rev= input.reverse();
   let value = rev[n-1];
   if(value){
     return value;
   }
   else
    return -1;
}