LeetCode Add Two Numbers

题目描述

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.

这个题其实是大数相加的链表实现，而且相比大数相加更加好处理一些，因为输入的数字是逆序的，因此在进行运算的时候可以从第一位开始相加
代码如下，有更好的方法欢迎指导交流

/**
 * 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 *p1 = l1, *p2 = l2, *p, *h, *node;
        h = NULL;
        int n = 0;
        while(p1 != NULL && p2 != NULL)  //l1,l2任何一方被处理完
        {
            node = new ListNode(n);
            n = node->val + p1->val + p2->val;
            node->val = n % 10;
            n /= 10;
            if(h == NULL)    //将头节点保存到h中,p链表中的最后一个节点
                h = p = node;
            else
            {
                p->next = node;
                p = node;
            }
            p1 = p1->next;
            p2 = p2->next;
        }
        while(p1 != NULL)  //l1链表中剩余的还有
        {
            node = new ListNode(n);
            n = node->val + p1->val;
            node->val = n % 10;
            n /= 10;
            p->next = node;
            p = node;
            p1 = p1->next;
        }
        while(p2 != NULL)  //l2链表中剩余的还有
        {
            node = new ListNode(n);
            n = node->val + p2->val;
            node->val = n % 10;
            n /= 10;
            p->next = node;
            p = node;
            p2 = p2->next;
        }
        if(n > 0)   //相加完还有进位，此处应当重视
        {
            node = new ListNode(n);
            p->next = node;
        }
        return h;
    }
};