Let's solve daily

let's go ahead and solve this together

i want you to take 2 linkedlists as inputs

example: [2, 5 , 8] + [1 , 0 , 1] returns [ 9, 5, 3]

reverse both of them and add them
return the reversed linked list

in java

Attached: rand.jpg (1426x1078, 179.91K)

(reverse (map #'+ '(2 5 8) '(9 5 3)))
>in java
no

alright alright i'll start us with a Node class..

public class Node (){
Node nextNode;
int value;
public Node(Node nextNode, int value){
this.nextNode = nextNode;
this value = value;
}}

create two strings
read each int from the lists, append to respective string
convert each string to Long or BigInteger
add
create linkedlist from resulting number and return

main runner class...

public static void main (String[] args){
Node input1 = new Node(2, new Node(5, new Node(8, null));
Node input2 = new Node(1, new Node(0, new Node(1, null));
}

we're gunna need a method that takes an input and reverses them, yo...

who wants to try for this

public Node reverse (Node node){
Node reversed = node;
//reverse me pls
return reversed;
}

Attached: 1643828001962.jpg (851x967, 165.51K)

alright let's modify our Node class just a little to add a method that points to the next node..

public class Node (){
Node nextNode;
int value;
public Node(Node nextNode, int value){
this.nextNode = nextNode;
this.value = value;
}
public Node next(){
return this.nextNode;
}
}

Baby's first leetcode

In Haskell, this is just:
type Digit = Int -- 0 to 9
type Carry = Int -- 0 or 1

addRev :: [Digit] -> [Digit] -> Carry -> [Digit] -> [Digit]
addRev (d1 : n1) (d2 : n2) c acc =
let s = d1 + d2 + c in
addRev n1 n2 (s `div` 10) (s `mod` 10 : acc)
addRev (d1 : n1) [] c acc = addRev (d1 : n1) [0] c acc
addRev [] (d2 : n2) c acc = addRev [0] (d2 : n2) c acc
addRev [] [] 0 acc = acc
addRev [] [] 1 acc = (1 : acc)

homework :: [Digit] -> [Digit] -> [Digit]
homework n1 n2 = add n1 n2 0 []

main = print $ homework [2, 5, 8] [1, 0, 1] -- [9, 5, 3]

homework n1 n2 = addRev n1 n2 0 []