Introduction
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 contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Solution 1: Direct Calculation with full adder
The idea is pretty simple: since the order is allready reversed, the lowest bit is automatically matched, using a Cin bit to preserve carray.
Pay attention that it’s possible the two number has different length, so in the while loop we need to consider 3 situation:
- l1 ended, l2 continue
- l2 ended, l1 continue
- l1,l2 both continue
Code:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
int c = 0;
ListNode* head = new ListNode(0);
ListNode* cnt = head;
while(l1 || l2)
{
if(!l1)
{
cnt->val = (l2->val + c)%10;
c = (l2->val + c)/10;
l2 = l2->next;
}
else if(!l2)
{
cnt->val = (l1->val + c)%10;
c = (l1->val + c)/10;
l1 = l1->next;
}
else
{
cnt->val = (l1->val + l2->val + c)%10;
c = (l1->val + l2->val + c)/10;
l1 = l1->next;
l2 = l2->next;
}
if(l1 || l2){
cnt->next = new ListNode(0);
cnt = cnt->next;
}
}
if(c!=0) cnt->next = new ListNode(c);
return head;
}