LeetCode 2. Add Tow Numbers

题目:

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

简单的加法模拟, 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) {
        //如果有一个链表为空,则直接返回另一个
        if(l1 == nullptr)
            return l2;
        if(l2 == nullptr)
            return l1;
            
        ListNode *p1 = l1, *p2 = l2;
        while(p1->next != nullptr || p2->next != nullptr){
            //填充较短的数, 高位添加0, 这一步可以用计算后再处理的办法替代
            if(p1->next == nullptr)
                p1->next = new ListNode(0);
            if(p2->next == nullptr)
                p2->next = new ListNode(0);
            p1 = p1->next, p2 = p2->next;
        }
        
        ListNode *head = new ListNode(0), *p = head;
        int jw = 0;//jw保存上一位计算后的进位
        for(p1 = l1, p2 = l2; p1 != nullptr && p2 != nullptr; p1 = p1->next, p2 = p2->next){
            //模拟每一位的加法
            int t = p1->val + p2->val + jw;
            jw = (t > 9 ? 1 : 0);
            
            t = t % 10;
            p->next = new ListNode(t);
            p = p->next;
        }
        if(jw == 1){
            //最后有进位的处理
            p->next = new ListNode(1);
        }
        
        return head->next;
    }
};