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
思路：用两个指针pa,pb同时遍历两个链表，如若其中一个指针为空，赋值为0,直到两个指针都为空，遍历结束。
/**
* 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) {
ListNode sum(-1);
int carry=0;
ListNode *prev=∑
ListNode *pa=l1,*pb=l2;
while(pa!=NULL||pb!=NULL)
{
const int ai=(pa==NULL?0:pa->val);
const int bi=(pb==NULL?0:pb->val);
const int value=(ai+bi+carry)%10;
carry=(ai+bi+carry)/10;
prev->next=new ListNode(value);
pa=(pa==NULL?NULL:pa->next);
pb=(pb==NULL?NULL:pb->next);
prev=prev->next;
}
if(carry>0)
prev->next=new ListNode(carry);
return sum.next;
}
};
原文：http://www.cnblogs.com/xiaoying1245970347/p/4576365.html