You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output:

思路非常简单,利用两个指针分别遍历两个链表,并且用一个变量表示是否有进位。某个链表遍历结束之后再将另一个链表连接在结果链表之后即可,若最后有进位需要添加一位。


/** 
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
int carry = 0;
ListNode* tail = new ListNode(0);
ListNode* ptr = tail;

while(l1 != NULL || l2 != NULL){
int val1 = 0;
if(l1 != NULL){
val1 = l1->val;
l1 = l1->next;
}

int val2 = 0;
if(l2 != NULL){
val2 = l2->val;
l2 = l2->next;
}

int tmp = val1 + val2 + carry;
ptr->next = new ListNode(tmp % 10);
carry = tmp / 10;
ptr = ptr->next;
}

if(carry == 1){
ptr->next = new ListNode(1);
}
return tail->next;
}
};