notes/OJ notes/pages/Leetcode Add-Two-Numbers.md

Leetcode Add-Two-Numbers

2022-09-08 15:22

Algorithms:

#algorithm #math

Data structures:

#DS #linked_list

Difficulty:

#coding_problem #difficulty_medium

Additional tags:

#leetcode

Revisions:

N/A

Links:

• Link to problem

---

Problem

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.

Examples

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.

Thoughts

[!summary] This is a elementary #math problem. For a list of similar questions, visit Elementary Math Problems

Done by using elementary math.

Use only one variable tmp to keep track of sum of 2 digits and carries.

Solution

/**
 * 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 {
  ListNode *appendNumber(ListNode *ptr, int num) {
    ListNode *ans = new ListNode(num, nullptr);
    ptr->next = ans;

    return ans;
  }

public:
  ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
    ListNode *ptr1 = l1, *ptr2 = l2;

    // the minimum size is 1, so we can safely do this.
    int tmp = ptr1->val + ptr2->val;

    ListNode *ans = new ListNode(tmp % 10);

    ListNode *tail = ans;

    tmp /= 10;
    ptr1 = ptr1->next;
    ptr2 = ptr2->next;

    while (ptr1 || ptr2 || tmp) {
      if (ptr1) {
        tmp += ptr1->val;
        ptr1 = ptr1->next;
      }

      if (ptr2) {
        tmp += ptr2->val;
        ptr2 = ptr2->next;
      }

      tail = appendNumber(tail, tmp % 10);

      tmp /= 10;
    }

    return ans;
  }
};