Add Two Numbers

描述

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: 7 -> 0 -> 8

【题意】

给你两个链表，表示两个非负整数。数字在链表中按反序存储，例如342在链表中为2->4->3。链表每一个节点包含一个数字（0-9）。

计算这两个数字和并以链表形式返回。

思路：

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;
/**
* 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;
}
};