Last updated on July 18th, 2024 at 04:51 am

Here, We see Add Two Numbers II LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Topics

Linked-List

Companies

Bloomberg, Microsoft

Level of Question

Medium

Add Two Numbers II LeetCode Solution

Problem Statement

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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 = [7,2,4,3], l2 = [5,6,4]
Output: [7,8,0,7]

Example 2:
Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [8,0,7]

Example 3:
Input: l1 = [0], l2 = [0]
Output: [0]

1. Add Two Numbers II LeetCode Solution C++

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* prev = NULL;
        while(head) {
            ListNode* nxt = head->next;
            head->next = prev;
            prev = head;
            head = nxt;
        }
        return prev;
    }

    ListNode* Helper(ListNode* l1, ListNode* l2) {
        ListNode* dummyHead = new ListNode(0);
        ListNode* tail = dummyHead;
        int carry = 0;
        while (l1 != nullptr || l2 != nullptr || carry != 0) {
            int digit1 = (l1 != nullptr) ? l1->val : 0;
            int digit2 = (l2 != nullptr) ? l2->val : 0;
            int sum = digit1 + digit2 + carry;
            int digit = sum % 10;
            carry = sum / 10;
            ListNode* newNode = new ListNode(digit);
            tail->next = newNode;
            tail = tail->next;
            l1 = (l1 != nullptr) ? l1->next : nullptr;
            l2 = (l2 != nullptr) ? l2->next : nullptr;
        }
        ListNode* result = dummyHead->next;
        delete dummyHead;
        return result;
    }

    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        l1 = reverseList(l1);
        l2 = reverseList(l2);
        ListNode* ans = Helper(l1, l2);
        return reverseList(ans);
    }
};

2. Add Two Numbers II LeetCode Solution Java

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode prev = null;
        ListNode curr = head;
        while (curr != null) {
            ListNode next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }

    public ListNode helper(ListNode l1, ListNode l2) {
        ListNode dummyHead = new ListNode(0);
        ListNode tail = dummyHead;
        int carry = 0;
        while (l1 != null || l2 != null || carry != 0) {
            int digit1 = (l1 != null) ? l1.val : 0;
            int digit2 = (l2 != null) ? l2.val : 0;
            int sum = digit1 + digit2 + carry;
            int digit = sum % 10;
            carry = sum / 10;
            ListNode newNode = new ListNode(digit);
            tail.next = newNode;
            tail = tail.next;
            l1 = (l1 != null) ? l1.next : null;
            l2 = (l2 != null) ? l2.next : null;
        }
        ListNode result = dummyHead.next;
        return result;
    }

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        l1 = reverseList(l1);
        l2 = reverseList(l2);
        ListNode ans = helper(l1, l2);
        return reverseList(ans);
    }
}

3. Add Two Numbers II Solution JavaScript

var addTwoNumbers = function(l1, l2) {
    let stack1 = [];
    let stack2 = [];
    while(l1) {
        stack1.push(l1.val);
        l1 = l1.next;
    }
    while(l2) {
        stack2.push(l2.val);
        l2 = l2.next;
    }
    let l3 = new ListNode(0);
    while(stack1.length || stack2.length) {
        let sum = 0;
        if(stack1.length) sum += stack1.pop();
        if(stack2.length) sum += stack2.pop();
        sum += l3.val;
        l3.val = sum%10;
        let head = new ListNode(Math.floor(sum/10));
        head.next = l3;
        l3 = head;
    }
    return (l3.val === 0) ? l3.next : l3;
};

4. Add Two Numbers II Solution Python

class Solution(object):
    def reverseList(self, head):
        prev = None
        curr = head
        while curr:
            nxt = curr.next
            curr.next = prev
            prev = curr
            curr = nxt
        return prev

    def helper(self, l1, l2):
        dummyHead = ListNode(0)
        tail = dummyHead
        carry = 0
        while l1 or l2 or carry:
            digit1 = l1.val if l1 else 0
            digit2 = l2.val if l2 else 0
            total = digit1 + digit2 + carry
            digit = total % 10
            carry = total // 10
            newNode = ListNode(digit)
            tail.next = newNode
            tail = tail.next
            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None
        result = dummyHead.next
        return result

    def addTwoNumbers(self, l1, l2):
        l1 = self.reverseList(l1)
        l2 = self.reverseList(l2)
        ans = self.helper(l1, l2)
        return self.reverseList(ans)