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
题目的含义是给两个链表,链表中每个节点的元素是各位的数字,从链表头开始作为低位,两个链表相加,最终的结果也存放在一个同样结构的链表中。具体解题思路为:首先生成一个result链表指针作为返回结果指向链表头,然后使用一个链表指针temp指向result的内容,然后用temp动态生成结果。整个处理过程仍然和大数处理相似,对应为相加,注意进位,但是此时没有先对链表遍历获取长度,因此可能会存在一个链表已经遍历完毕而另一个还未完全完毕的情况。所以需要对while(l1 !=NULL && l2 !=NULL)循环外的l1和l2检查是否已经到链表尾,未到链表尾的需要再和进位位相加生成节点在结果链表中。
/**
* 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) {
if(l1==NULL&&l2==NULL) return NULL;
if(l1==NULL&&l2 != NULL) return l2;
if(l1!=NULL&&l2 == NULL) return l1;
ListNode *result = new ListNode(0);
ListNode *temp = result;
ListNode *q =NULL;
int flag = 0;
while(l1 !=NULL && l2 !=NULL)
{
int total = flag + l1->val+l2->val;
temp->val = total%10;
temp->next = new ListNode(0);
q = temp;
temp = temp->next;
flag = total/10;
l1 = l1->next;
l2 = l2->next;
}
if(l1==NULL&&l2==NULL)
{
if(!flag)
q->next = NULL;
else
temp->val = flag;
}
if(l1 != NULL)
{
while(l1 !=NULL)
{
int total = flag +l1->val;
temp->val = total%10;
temp->next = new ListNode(0);
q = temp;
temp = temp->next;
flag = total/10;
l1 = l1->next;
}
if(!flag)
q->next = NULL;
else
temp->val = flag;
}
if(l2 != NULL)
{
while(l2 !=NULL)
{
int total = flag +l2->val;
temp->val = total%10;
temp->next = new ListNode(0);
q = temp;
temp = temp->next;
flag = total/10;
l2 = l2->next;
}
if(!flag)
q->next = NULL;
else
temp->val = flag;
}
return result;
}