Add Two Numbers

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：将两个链表从表头开始逐个相加，遇到大于10的则向后一节点进1。使用两个指针分别指向两个链表，且进位值用mod记录，如果一个链表没有结束，则将这个链表的剩余部分添加进去，注意进位。如果都结束了，但最后一位出现了进位，则要添加一个结点到新链表中。

/**
 * 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)
            return l2;
        else if(l2==NULL)
            return l1;
        ListNode *lHead=NULL;
        ListNode *pList=NULL;
        int mod=0;
        while(l1!=NULL&&l2!=NULL)
        {
            int temp=l1->val+l2->val+mod;
            mod=temp/10;
            temp%=10;
            if(lHead==NULL)
            {
                lHead=new ListNode(temp);
                pList=lHead;
            }
            else
            {
                pList->next=new ListNode(temp);
                pList=pList->next;
            }
            l1=l1->next;
            l2=l2->next;
        }
        while(l1!=NULL)
        {
            int temp=l1->val+mod;
            mod=temp/10;
            temp%=10;
            if(lHead==NULL)
            {
                lHead=new ListNode(temp);
                pList=lHead;
            }
            else
            {
                pList->next=new ListNode(temp);
                pList=pList->next;
            }
            l1=l1->next;
        }
        while(l2!=NULL)
        {
            int temp=l2->val+mod;
            mod=temp/10;
            temp%=10;
            if(lHead==NULL)
            {
                lHead=new ListNode(temp);
                pList=lHead;
            }
            else
            {
                pList->next=new ListNode(temp);
                pList=pList->next;
            }
            l2=l2->next;
        }
        if(mod!=0)//注意这里。
            pList->next=new ListNode(mod);
        return lHead;
    }
};

 借用网上大神的代码比较简洁，但是思路差不多。

/**
 * 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)
            return l2;
        else if(l2==NULL)
            return l1;
        ListNode dummy(0);
        ListNode *pHead=&dummy;
        int carry=0;
        while(l1!=NULL||l2!=NULL||carry!=0)
        {
            int a=0,b=0;
            if(l1!=NULL)
            {
                a=l1->val;
                l1=l1->next;
            }
            if(l2!=NULL)
            {
                b=l2->val;
                l2=l2->next;
            }
            int sum=a+b+carry;
            carry=sum/10;
            sum%=10;
            pHead->next=new ListNode(sum);
            pHead=pHead->next;
        }
        return dummy.next;
    }
};