2. Add Two Numbers
Question：
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.
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 {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode *curcor = new ListNode(0);
ListNode *temp = curcor;
int carry = 0;
while(l1 != NULL || l2 != NULL)
{
int num = 0;
if(l1 != NULL && l2 != NULL)
{
num = l1->val+l2->val;
l2 = l2->next;
l1 = l1->next;
}else if(l1 == NULL && l2 != NULL)
{
num = l2->val;
l2 = l2->next;
}else if(l1 != NULL && l2 == NULL)
{
num = l1->val;
l1 = l1->next;
}
int current = (num+carry)%10;
ListNode *p = new ListNode(current);
temp->next =p;
temp = p;
carry = (num+carry)/10;
if(l1 == NULL && l2 == NULL && carry != 0)
{
ListNode *p = new ListNode(carry);
temp->next =p;
temp = p;
}
}
return curcor->next;
}
};
