问题描写叙述:

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

解决方式：

/** * 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;        int c = 0;        ListNode* result = new ListNode(0);        ListNode* pNode = result;        pNode->next = NULL;        pNode->val = (l1->val+l2->val)%10;        c = (l1->val+l2->val)/10;        l1 = l1->next;        l2 = l2->next;        while(l1!=NULL&&l2!=NULL)        {            ListNode* temp = new ListNode(0);            temp->next = NULL;            temp->val = (l1->val+l2->val+c)%10;            c = (l1->val+l2->val+c)/10;            pNode->next = temp;            pNode = pNode->next;            l1 = l1->next;            l2 = l2->next;        }        while(l1!=NULL)        {            ListNode* temp = new ListNode(0);            temp->next = NULL;            temp->val = (l1->val+c)%10;            c = (l1->val+c)/10;            pNode->next = temp;            pNode = pNode->next;            l1=l1->next;        }         while(l2!=NULL)        {            ListNode* temp = new ListNode(0);            temp->next = NULL;            temp->val = (l2->val+c)%10;            c = (l2->val+c)/10;            pNode->next = temp;            pNode = pNode->next;            l2=l2->next;        }        if(c)        {            ListNode* temp = new ListNode(c);            pNode->next = temp;        }        return result;    }};