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;

    }

};