Leetcode OJ: Add Two Numbers

Add Two Numbers

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

大非负整数相加，注意数字是怎么表示的就好，链表头是低位，尾是高位，另外最后一位记得处理，也不要忘了0的情况。

/**
 * 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;
        ListNode* ret = NULL;
        ListNode* p = ret;
        // 逐位相加
        while (p1 != NULL && p2 != NULL) {
            ListNode* tmp = new ListNode(p1->val + p2->val);
            // 判断是不是链表头
            if (ret == NULL) {
                ret = tmp;
                p = ret;
            } else {
                p->next = tmp;
                p = tmp;
            }
            p1 = p1->next;
            p2 = p2->next;
        }
        
        if (p1 != NULL) {
            p->next = p1;
        }
        if (p2 != NULL) {
            p->next = p2;
        }
        
        // 为空时要返回0
        if (ret == NULL)
            return new ListNode(0);
        
        // 处理进位问题
        p = ret;
        while (p->next != NULL) {
            p->next->val += p->val / 10;
            p->val = p->val % 10;
            p = p->next;
        }
        // 处理最高位
        if (p->val >= 10) {
            ListNode* tmp = new ListNode(p->val / 10);
            p->next = tmp;
            p->val = p->val % 10;
        }
        
        return ret;
        
    }
};