2 Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

本题考查列表的基本操作，采用一个头指针和一个尾指针控制列表的插入和读取，需要注意的是计算两数加和时对进位项的处理。第一次提交的代码如下：

/**
 * 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) {
        
        ListNode* head = NULL;
        ListNode* tail = NULL;
        ListNode* p = l1;
        ListNode* q = l2;
        int carry = 0;
        
        while(p != NULL || q != NULL)
        {
            
            int x = (p != NULL) ? p->val: 0;
            int y = (q != NULL) ? q->val: 0;
            int sum = x + y + carry;
            carry = sum/10;
            
            if(head == NULL) 
            {
                head = new ListNode(sum%10);
                tail = head;
            }
            else {
                tail->next = new ListNode(sum%10);
                tail = tail->next;
            }
            if(p != NULL) p = p->next;
            if(q != NULL) q = q->next;
            
        }
        
        if(carry > 0)
        {
            tail->next = new ListNode(carry);
            tail = tail->next;
        }
        
        return head;
    }
};

第一次提交代码后发现运算速度并不快（54.2%），内存也不是最优，于是进行第二次迭代。

第二次迭代优化的内容有：

1. 省掉了部分参数
2. 合并了部分条件判断
3. 简化了条件判断语句

最后的运行速度达到20ms(92.87%), 内存少于（71.69%）

/**
 * 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) {
        
        ListNode* head = NULL;
        ListNode* tail = NULL;
        int carry = 0;
        
        while(l1||l2)
        {
            int x,y,sum;
            
            if(l1)
            {
                x  = l1->val;
                l1 = l1->next;
            }
            else x = 0;
            if(l2)
            {
                y = l2->val;
                l2 = l2->next;
            }
            else y = 0;

            sum = x + y + carry;
            carry = sum/10;
            
            if(head == NULL) 
            {
                head = new ListNode(sum%10);
                tail = head;
            }
            else {
                tail->next = new ListNode(sum%10);
                tail = tail->next;
            }
            
        }
        
        if(carry > 0)
        {
            tail->next = new ListNode(carry);
            tail = tail->next;
        }
        
        return head;
    }
};