2. Add Two Numbers

Description

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example 1:

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

Example 2:

Input: l1 = [0], l2 = [0]
Output: [0]

Example 3:

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]

Constraints:

The number of nodes in each linked list is in the range [1, 100].
0 <= Node.val <= 9
It is guaranteed that the list represents a number that does not have leading zeros.

Solutions

Solution 1: Simulation

We traverse two linked lists $l_1$ and $l_2$ at the same time, and use the variable $carry$ to indicate whether there is a carry.

Each time we traverse, we take out the current bit of the corresponding linked list, calculate the sum with the carry $carry$, and then update the value of the carry. Then we add the current bit to the answer linked list. If both linked lists are traversed, and the carry is $0$, the traversal ends.

Finally, we return the head node of the answer linked list.

The time complexity is $O(\max (m, n))$, where $m$ and $n$ are the lengths of the two linked lists. We need to traverse the entire position of the two linked lists, and each position only needs $O(1)$ time. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        int carry = 0;
        ListNode cur = dummy;
        while (l1 != null || l2 != null || carry != 0) {
            int s = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
            carry = s / 10;
            cur.next = new ListNode(s % 10);
            cur = cur.next;
            l1 = l1 == null ? null : l1.next;
            l2 = l2 == null ? null : l2.next;
        }
        return dummy.next;
    }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* dummy = new ListNode();
        int carry = 0;
        ListNode* cur = dummy;
        while (l1 || l2 || carry) {
            int s = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + carry;
            carry = s / 10;
            cur->next = new ListNode(s % 10);
            cur = cur->next;
            l1 = l1 ? l1->next : nullptr;
            l2 = l2 ? l2->next : nullptr;
        }
        return dummy->next;
    }
};

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def addTwoNumbers(
        self, l1: Optional[ListNode], l2: Optional[ListNode]
    ) -> Optional[ListNode]:
        dummy = ListNode()
        carry, curr = 0, dummy
        while l1 or l2 or carry:
            s = (l1.val if l1 else 0) + (l2.val if l2 else 0) + carry
            carry, val = divmod(s, 10)
            curr.next = ListNode(val)
            curr = curr.next
            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None
        return dummy.next

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
	dummy := &ListNode{}
	carry := 0
	cur := dummy
	for l1 != nil || l2 != nil || carry != 0 {
		s := carry
		if l1 != nil {
			s += l1.Val
		}
		if l2 != nil {
			s += l2.Val
		}
		carry = s / 10
		cur.Next = &ListNode{s % 10, nil}
		cur = cur.Next
		if l1 != nil {
			l1 = l1.Next
		}
		if l2 != nil {
			l2 = l2.Next
		}
	}
	return dummy.Next
}

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function addTwoNumbers(l1: ListNode | null, l2: ListNode | null): ListNode | null {
    const dummy = new ListNode();
    let cur = dummy;
    let sum = 0;
    while (sum !== 0 || l1 != null || l2 != null) {
        if (l1 != null) {
            sum += l1.val;
            l1 = l1.next;
        }
        if (l2 != null) {
            sum += l2.val;
            l2 = l2.next;
        }
        cur.next = new ListNode(sum % 10);
        cur = cur.next;
        sum = Math.floor(sum / 10);
    }
    return dummy.next;
}

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function (l1, l2) {
    const dummy = new ListNode();
    let carry = 0;
    let cur = dummy;
    while (l1 || l2 || carry) {
        const s = (l1?.val || 0) + (l2?.val || 0) + carry;
        carry = Math.floor(s / 10);
        cur.next = new ListNode(s % 10);
        cur = cur.next;
        l1 = l1?.next;
        l2 = l2?.next;
    }
    return dummy.next;
};

/**
 * Definition for a singly-linked list.
 * class ListNode {
 *     public $val = 0;
 *     public $next = null;
 *     function __construct($val = 0, $next = null) {
 *         $this->val = $val;
 *         $this->next = $next;
 *     }
 * }
 */
class Solution {
    /**
     * @param ListNode $l1
     * @param ListNode $l2
     * @return ListNode
     */
    function addTwoNumbers($l1, $l2) {
        $dummy = new ListNode(0);
        $current = $dummy;
        $carry = 0;

        while ($l1 !== null || $l2 !== null) {
            $x = $l1 !== null ? $l1->val : 0;
            $y = $l2 !== null ? $l2->val : 0;

            $sum = $x + $y + $carry;
            $carry = (int) ($sum / 10);
            $current->next = new ListNode($sum % 10);
            $current = $current->next;

            if ($l1 !== null) {
                $l1 = $l1->next;
            }

            if ($l2 !== null) {
                $l2 = $l2->next;
            }
        }

        if ($carry > 0) {
            $current->next = new ListNode($carry);
        }

        return $dummy->next;
    }
}

# Definition for singly-linked list.
# class ListNode
#     attr_accessor :val, :next
#     def initialize(val = 0, _next = nil)
#         @val = val
#         @next = _next
#     end
# end
# @param {ListNode} l1
# @param {ListNode} l2
# @return {ListNode}
def add_two_numbers(l1, l2)
  dummy = ListNode.new()
  carry = 0
  cur = dummy
  while !l1.nil? || !l2.nil? || carry > 0
      s = (l1.nil? ? 0 : l1.val) + (l2.nil? ? 0 : l2.val) + carry
      carry = s / 10
      cur.next = ListNode.new(s % 10)
      cur = cur.next
      l1 = l1.nil? ? l1 : l1.next
      l2 = l2.nil? ? l2 : l2.next
  end
  dummy.next
end

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode AddTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode();
        int carry = 0;
        ListNode cur = dummy;
        while (l1 != null || l2 != null || carry != 0) {
            int s = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
            carry = s / 10;
            cur.next = new ListNode(s % 10);
            cur = cur.next;
            l1 = l1 == null ? null : l1.next;
            l2 = l2 == null ? null : l2.next;
        }
        return dummy.next;
    }
}

import std/[strutils, algorithm]

type
  Node[int] = ref object
    value: int
    next: Node[int]

  SinglyLinkedList[T] = object
    head, tail: Node[T]

proc append[T](list: var SinglyLinkedList[T], data: T = nil): void =
  var node = Node[T](value: data)
  if list.head.isNil:
    list.head = node
    list.tail = node
  else:
    list.tail.next = node
    list.tail = node

proc preview[T](list: SinglyLinkedList[T]): string =
  var s: seq[T]
  var n = list.head
  while not n.isNil:
    s.add n.value
    n = n.next
  result = s.join(" -> ")

proc addTwoNumbers(l1: var SinglyLinkedList, l2: var SinglyLinkedList): SinglyLinkedList[int] =
  var
    aggregate: SinglyLinkedList
    psum: seq[char]
    temp_la, temp_lb: seq[int]

  while not l1.head.isNil:
    temp_la.add(l1.head.value)
    l1.head = l1.head.next

  while not l2.head.isNil:
    temp_lb.add(l2.head.value)
    l2.head = l2.head.next

  psum = reversed($(reversed(temp_la).join("").parseInt() + reversed(temp_lb).join("").parseInt()))
  for i in psum: aggregate.append(($i).parseInt())

  result = aggregate

var list1: SinglyLinkedList[int]
var list2: SinglyLinkedList[int]

for i in @[2, 4, 3]: list1.append(i)
for i in @[5, 6, 4]: list2.append(i)

echo(preview(list1))
echo(preview(list2))
echo(preview(addTwoNumbers(list1, list2)))

// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn add_two_numbers(
        mut l1: Option<Box<ListNode>>,
        mut l2: Option<Box<ListNode>>
    ) -> Option<Box<ListNode>> {
        let mut dummy = Some(Box::new(ListNode::new(0)));
        let mut cur = &mut dummy;
        let mut sum = 0;
        while l1.is_some() || l2.is_some() || sum != 0 {
            if let Some(node) = l1 {
                sum += node.val;
                l1 = node.next;
            }
            if let Some(node) = l2 {
                sum += node.val;
                l2 = node.next;
            }
            cur.as_mut().unwrap().next = Some(Box::new(ListNode::new(sum % 10)));
            cur = &mut cur.as_mut().unwrap().next;
            sum /= 10;
        }
        dummy.unwrap().next.take()
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
class Solution {
    func addTwoNumbers(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        var dummy = ListNode.init()
        var carry = 0
        var l1 = l1
        var l2 = l2
        var cur = dummy
        while l1 != nil || l2 != nil || carry != 0 {
            let s = (l1?.val ?? 0) + (l2?.val ?? 0) + carry
            carry = s / 10
            cur.next = ListNode.init(s % 10)
            cur = cur.next!
            l1 = l1?.next
            l2 = l2?.next
        }
        return dummy.next
    }
}

2 - Add Two Numbers

2. Add Two Numbers

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2. Add Two Numbers

Description

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All Problems

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