LeetCode 2_Add Two Numbers

LeetCode 2_Add Two Numbers

题目描述：

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

题目很清晰（虽然我看了半天），无非链表相加！

注意：

1、链表长度没说一样长。

2、考虑最后一个节点有可能进位，此时需要新建节点。

首先不加思索写出的代码：

 ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)    {        ListNode* l3 = l1;        int tmp1 = 0, tmp = 0;        while(l1->next != NULL && l2->next != NULL)        {            tmp1 = l1->val;            l1->val = (l1->val + l2->val + tmp) % 10;            tmp = (tmp1 + l2->val + tmp) / 10;            l1 = l1->next;            l2 = l2->next;        }        if(l1->next == NULL && l2->next == NULL)        {            tmp1 = l1->val;            l1->val = (l1->val + l2->val + tmp) % 10;            tmp = (tmp1 + l2->val + tmp) / 10;            if(tmp == 1)                l1->next = new ListNode(1);            return l3;        }        if(l1->next != NULL)        {            tmp1 = l1->val;            l1->val = (l1->val + l2->val + tmp) % 10;            tmp = (tmp1 + l2->val + tmp) / 10;            l1 = l1->next;            while(l1->next != NULL)            {                if(l1->val + tmp < 10)                {                    l1->val = l1->val + tmp;                    return l3;                }                else                {                    l1->val = (l1->val + tmp) % 10;                    tmp = 1;                    l1 = l1->next;                }            }            if(l1->val + tmp >= 10)            {                l1->val = (l1->val + tmp) % 10;                l1->next = new ListNode(1);                return l3;            }            else            {                l1->val = (l1->val + tmp) % 10;                return l3;            }        }        if(l2->next != NULL)        {            tmp1 = l1->val;            l1->val = (l1->val + l2->val + tmp) % 10;            tmp = (tmp1 + l2->val + tmp) / 10;            l1->next = l2->next;            l1 = l1->next;            while(l1->next != NULL)            {                if(l1->val + tmp < 10)                {                    l1->val = l1->val + tmp;                    return l3;                }                else                {                    l1->val = (l1->val + tmp) % 10;                    tmp = 1;                    l1 = l1->next;                }            }            if(l1->val + tmp >= 10)            {                l1->val = (l1->val + tmp) % 10;                l1->next = new ListNode(1);                return l3;            }            else            {                l1->val = (l1->val + tmp) % 10;                return l3;            }        }    }

这代码，看起来好low，首先代码重复性太大，从代码的编写就可以有大量的优化空间。

最终优化为（其实也没什么优化，只是删掉了一些重复）：

 ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)    {        ListNode* l3 = l1;        int tmp1 = 0, tmp = 0;        while(l1->next != NULL && l2->next != NULL)        {            tmp1 = l1->val;            l1->val = (l1->val + l2->val + tmp) % 10;            tmp = (tmp1 + l2->val + tmp) / 10;            l1 = l1->next;            l2 = l2->next;        }        tmp1 = l1->val;        l1->val = (l1->val + l2->val + tmp) % 10;        tmp = (tmp1 + l2->val + tmp) / 10;        if(l1->next == NULL && l2->next == NULL)        {            if(tmp == 1)                l1->next = new ListNode(1);            return l3;        }        else        {            if(l2->next != NULL)                l1->next = l2->next;            l1 = l1->next;            while(l1->next != NULL)            {                if(l1->val + tmp < 10)                {                    l1->val = l1->val + tmp;                    return l3;                }                else                {                    l1->val = (l1->val + tmp) % 10;                    tmp = 1;                    l1 = l1->next;                }            }            if(l1->val + tmp >= 10)            {                l1->val = (l1->val + tmp) % 10;                l1->next = new ListNode(1);                return l3;            }            else            {                l1->val = (l1->val + tmp) % 10;                return l3;            }        }</span>

其实有一种更简单的方法是再新建一个链表，不过就是会浪费空间

 ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)    {        int addNum = 0;        ListNode *l3 = new ListNode(0);        ListNode *ptr = l3;        while(l1 != NULL || l2 != NULL)        {            int val1 = 0;              if(l1 != NULL)            {                  val1 = l1->val;                  l1 = l1->next;              }                            int val2 = 0;              if(l2 != NULL)            {                  val2 = l2->val;                  l2 = l2->next;              }                          ListNode *temp = new ListNode((val1 + val2 + addNum) % 10);            ptr->next = temp;            ptr = temp;            addNum = (val1 + val2 + addNum) / 10;        }        if(addNum == 1)           ptr->next = new ListNode(1);        return l3->next;   }