OJ 002 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
首先想到的是用递归来完成这个问题。
代码：

class Solution {public:    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {        ListNode *ret = NULL;        cout << "使用递归完成" << endl;        ret = addTwoPoint(l1, l2, 0);        return ret;    }    ListNode *addTwoPoint(ListNode *l1, ListNode *l2, int carry)    {        /***********************        参数：l1第一条链表              l2第二条链表              carry进位        ***********************/        if (l1 == NULL&&l2 == NULL)        {            if (carry == 0)                return NULL;            else{                ListNode *npoint = new ListNode(carry);                return npoint;            }        }        else if (l1 == NULL&&l2 != NULL)        {            int tmp = l2->val + carry;            ListNode *npoint = new ListNode(tmp % 10);            npoint->next = addTwoPoint(l1, l2->next, tmp / 10);            return npoint;        }        else if (l1 != NULL&&l2 == NULL)        {            int tmp = l1->val + carry;            ListNode *npoint = new ListNode(tmp % 10);            npoint->next = addTwoPoint(l1->next, l2, tmp / 10);            return npoint;        }        else        {            int tmp = l1->val + l2->val + carry;            ListNode *npoint = new ListNode(tmp % 10);            npoint->next = addTwoPoint(l1->next, l2->next, tmp / 10);            return npoint;        }    }};

上面是使用递归进行计算的代码，很简单，那如果改成循环做，该如何做呢？代码如下：

class Solution {public:    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {        /*        使用循环完成        */        //cout << "使用循环完成" << endl;        ListNode *ret = new ListNode(-1);        int carry = 0;        ListNode *listA = l1;        ListNode *listB = l2;        ListNode *tmp = ret;        int sum = 0;        while (listA != NULL&&listB != NULL)        {            sum = listA->val + listB->val + sum;            tmp->next = new ListNode(sum % 10);            sum = sum / 10;            listA = listA->next;            listB = listB->next;            tmp = tmp->next;        }        while (listA != NULL)        {            sum = listA->val  + sum;            tmp->next = new ListNode(sum % 10);            sum = sum / 10;            listA = listA->next;            tmp = tmp->next;        }        while (listB != NULL)        {            sum =  listB->val + sum;            tmp->next = new ListNode(sum % 10);            sum = sum / 10;            listB = listB->next;            tmp = tmp->next;        }        if (sum != 0)        {            tmp->next = new ListNode(sum % 10);        }        ret = ret->next;            //去掉第一个无用节点，如果严格来说，应该释放这块内存        return ret;    }};

假如我们修改一下题目，让最高位在链表的开头，然后要怎么做呢？
代码：

class Solution {public:    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {        int len1 = 0;        int len2 = 0;        ListNode *t1 = l1;        ListNode *t2 = l2;        while (t1 != NULL)        {            t1 = t1->next;            len1++;        }        while (t2 != NULL)        {            t2 = t2->next;            len2++;        }        if (len1 < len2)        {            return addTwoNumbers(l2, l1);        }        int step = len1 - len2;        t1 = l1;        t2 = l2;        ListNode *ret = new ListNode(0);        if (step == 0)        {            //            ret->next = t1;            int carry = addTwoPoint(t1, t2);            ret->val = carry;            if (carry == 0)                ret = ret->next;            return ret;        }        while (step > 1)        {            t1 = t1->next;            step--;        }        ret = l1;        int carry = addTwoPoint(t1->next, t2);        t1->val += carry;        return ret;    }    int addTwoPoint(ListNode *l1, ListNode *l2)    {        /***********************        参数：l1第一条链表        l2第二条链表        carry进位        ***********************/        int carry;        if (l1->next == NULL)        {            carry = 0;        }        else        {            carry = addTwoPoint(l1->next, l2->next);        }        int sum = l1->val + l2->val + carry;        l1->val = sum % 10;        return sum / 10;    }};