tags: Medium Link List Math Recursion

2. Add Two Numbers

Desciption

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

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.

Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    if (!l1)
        return l2;
    if (!l2)
        return l1;
    int carry = 0;
    struct ListNode *tmp, *ans, *last;
    ans = (struct ListNode*)malloc(sizeof(struct ListNode));
    tmp = ans;
    while (l1 != NULL || l2 != NULL || carry != 0) {
        tmp->val =
            ((l1 != NULL ? l1->val : 0) + (l2 != NULL ? l2->val : 0) + carry) % 10;
        carry =
            ((l1 != NULL ? l1->val : 0) + (l2 != NULL ? l2->val : 0) + carry) / 10;
        if (l1 != NULL)
            l1 = l1->next;
        if (l2 != NULL)
            l2 = l2->next;
        last = tmp;
        tmp->next = (struct ListNode*)malloc(sizeof(struct ListNode));
        tmp = tmp->next;
    }
    last->next = NULL;
    return ans;
}

Complexity

Space	Time
$O(Max(l1, l2))$	$O(Max(l1, l2))$

Result

• Runtime: 20 ms, 61.42 %
• Memory usage: 7.8 MB, 29.34 %