Leetcode 2. 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

分析：两个链表中包含的数不一定一样多！！

 

/**
 * 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) {
        int flag = 0, add;
        //创建一个节点要开辟空间，不能只定义一个指针，否则没有指向
        ListNode* head = new ListNode(0), *p, *q, *l3;
        p = l1;
        q = l2;
        l3 = head;
        while(p != NULL && q != NULL){
            l3->next = new ListNode((p->val + q->val + flag) % 10);
            flag = (p->val + q->val  + flag) / 10;
            l3 = l3->next;
            p = p->next;
            q = q->next;
        }
        while(p != NULL){
            l3->next = new ListNode((p->val + flag) % 10);
            flag = (p->val + flag) / 10;
            l3 = l3->next;
            p = p->next;
        }
         while(q != NULL){
            l3->next = new ListNode((q->val + flag) % 10);
            flag = (q->val + flag) / 10;
            l3 = l3->next;
            q = q->next;
        }
        if(flag == 1){//加入两个链表【9,9】，【8,9】，可知最后两个数加起来为19，需要在再创建一个节点最后放置1
            l3->next = new ListNode(flag);
            l3 = l3->next;
        }
        return head->next;
        
    }
};