LeetCode //C - 2. Add Two Numbers

2. Add Two Numbers

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.

From: LeetCode

Link: 2. Add Two Numbers

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Solution:

Ideas：

This function creates a dummy head for the result linked list, then iterates through both input lists l1 and l2. It adds corresponding digits along with any carry from the previous addition. If one list is longer than the other, it continues adding with the remaining digits. After the loop, if there is a carry left, it adds a new node with the carry value.

Code:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode *dummyHead = (struct ListNode*) malloc(sizeof(struct ListNode));
    struct ListNode *current = dummyHead;
    int carry = 0;
    
    while (l1 != NULL || l2 != NULL || carry != 0) {
        int sum = carry;
        if (l1 != NULL) {
            sum += l1->val;
            l1 = l1->next;
        }
        if (l2 != NULL) {
            sum += l2->val;
            l2 = l2->next;
        }
        carry = sum / 10;
        current->next = (struct ListNode*) malloc(sizeof(struct ListNode));
        current = current->next;
        current->val = sum % 10;
    }
    
    current->next = NULL; // Terminate the list
    struct ListNode *result = dummyHead->next; // The result is in the next node after dummy head
    free(dummyHead); // Free the dummy head node
    return result;
}