add_two_numbers.cpp

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
// 44 ms solution, O(max(m,n)) => space and time complexity both
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* l3_head = new ListNode(0);
        auto l3 = l3_head;
        int carry = 0;
        while (l1 != NULL and l2 != NULL){
            l3->next = new ListNode( l1->val + l2->val + carry );
            carry = l3->next->val / 10;
            l3->next->val -= carry * 10;
            l3 = l3->next;
            l1 = l1->next;
            l2 = l2->next;
        }
        
        ListNode* temp;
        if (l1 == NULL)
            temp = l2;
        else 
            temp = l1;
        
        while (temp != NULL){
            l3->next = new ListNode( temp->val + carry );
            carry = l3->next->val / 10;
            l3->next->val -= carry * 10;
            temp = temp->next;
            l3 = l3->next;
        }
        
        if (carry > 0){
            l3->next = new ListNode( carry );
        }
        
        l3_head = l3_head->next;
        
        return l3_head;
    }
};


// 16 ms solution - Credit: LeetCode
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */

auto desyncio = []()
{
    std::ios::sync_with_stdio(false);
    cin.tie(nullptr);
    return nullptr;
}();

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* root = NULL, *current = NULL;
        
        int c = 0;
        while(l1 || l2 || c) {
            int x = l1 ? l1->val : 0;
            int y = l2 ? l2->val : 0;
            int z = x + y + c;
            int r = z % 10;
            c = z/10;
            if(!current) {
                root = current = new ListNode(r);
            } else {
                current->next = new ListNode(r);
                current = current->next;
            }
            
            l1 = l1 ? l1->next: NULL;
            l2 = l2 ? l2->next: NULL;
        }
        
        return root;
    }
};