标签：
题目
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
分析：
这道题并不难，主要是细节的分析和控制，一个特殊情况就是两个列都遍历完之后，但是进位为1，此时要新建一个尾节点来存储这个数。
参考代码：
/**
* 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) {
bool jw=0;
int num=0;
int sum=0;
int count = 0;
ListNode *result, *pre_node, *cur_node;
while (l1 != NULL && l2 != NULL)
{
sum = l1->val + l2->val + jw;
num = sum % 10;
cur_node = (ListNode *)malloc(sizeof(ListNode));
cur_node->val = num;
cur_node->next = NULL;
if (count == 0)
{
result = cur_node;
pre_node = cur_node;
}
else
{
pre_node->next = cur_node;
pre_node=pre_node->next;
}
l1 = l1->next;
l2 = l2->next;
count++;
jw=sum/10;
}
while (l1 != NULL)
{
sum = l1->val+jw;
num = sum % 10;
l1->val=num;
cur_node->next = l1;
cur_node=cur_node->next;
l1 = l1->next;
jw=sum/10;
}
while (l2 != NULL)
{
sum = l2->val+jw;
num = sum % 10;
l2->val=num;
cur_node->next = l2;
cur_node=cur_node->next;
l2 = l2->next;
jw=sum/10;
}
if(jw==1)
{
ListNode *tail=(ListNode *)malloc(sizeof(ListNode));
tail->val=jw;
cur_node->next=tail;
cur_node=cur_node->next;
}
cur_node->next = NULL;
return result;
}
};
标签：
原文地址：http://www.cnblogs.com/yanqi0124/p/4244985.html
