leetcode第四题--Add Two Numbers

Problem:

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

一开始就是想到先把第一个list转换成一个数，然后把第二个转换成第二个数，然后相加后，再把相加的值变成list。代码如下

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        int a = 0, b = 0, ans = 0, flag = 1;
        int index1 = 0, index2 = 0;
        ListNode *temp1, *temp2, *result;
        temp1 = l1;
        temp2 = l2;
        while(temp1->next != NULL )
        {
            index1++;
            a += (temp1 -> next -> val) * (int)pow(10,index1);
        }
        while(temp2 -> next != NULL)
        {
            index2++;
            b += temp2 ->next -> val * (int)pow(10, index2);
        }
        ans = a + b;
        result -> val = ans%10;
        result -> next = NULL;
        flag = ans/10;
        while(flag)
        {
            ListNode *added = new ListNode(flag%10);
            result ->next = added;
            flag = flag/10;
        }
        return result;
    }

然后是Time Limit Exceed了。那就不能这样做，应该直接在链表相加。加到某个链表结束为止。要用中间变量记住当前的进位。如果最后进位不为零（也就是为1）的话，那还是需要记录的。代码贴出如下：

class Solution {  
public:  
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) 
    {  
        ListNode * ans = NULL, *last = NULL;  
        int up = 0;  
        while (l1 != NULL && l2 != NULL) 
        {  
            int tmp = l1->val + l2->val + up;  
            up = tmp / 10;  
            if (last == NULL) 
            {  
                ans = new ListNode(tmp % 10);  
                last = ans;  
            } 
            else  
                last = pushBack(last, tmp % 10);  
            l1 = l1->next;  
            l2 = l2->next;  
        }  
        while (l1 != NULL) 
        {  
            int tmp = l1->val + up;  
            last = pushBack(last, tmp % 10);  
            up = tmp / 10;  
            l1 = l1->next;  
        }  
        while (l2 != NULL) 
        {  
            int tmp = l2->val + up;  
            last = pushBack(last, tmp % 10);  
            up = tmp / 10;  
            l2 = l2->next;  
        }  
        if (up == 1) 
        {  
            ListNode * l = new ListNode(up);  
            last->next = l;  
        }  
        return ans;  
    }  
  
    ListNode * pushBack(ListNode * last, int val)
    {  
        ListNode * l = new ListNode(val);  
        last->next = l;  
        return l;  
    }  
};  

还是要感谢suool大神，改天一定要再做一次看看是不是真的掌握了。自己真的水平有限啊。不过只要肯努力，一天进步一点点就好。让cnblogs记录我的学习过程，come on！