LeetCode 2_Add Two Numbers

标签：leetcode   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

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;
   }

LeetCode 2_Add Two Numbers

标签：leetcode   add two numbers   

原文地址：http://blog.csdn.net/xwchao2014/article/details/47062823