[LeetCode]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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

题目本身不难，但是细节上容易出错。

因为数字是倒序存储在链表中，所以只需要逐位相加并记录进位即可。

当两个链表长度不等时，需要对较短的链表补零，例如(1 -> 2) + (1)需要处理成(1 -> 2) + (1 -> 0)。

当数字最高位即链表尾相加后有进位时，需要在表尾加一个1，例如(5) + (5)结果为(0 -> 1)。

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    int carry = 0;  //记录进位    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        ListNode *sum = NULL;        ListNode *head = NULL;        int temp = 0;        while (l1 != NULL || l2 != NULL) {                        /*对较短的链表补零*/            if (l1 == NULL) {        l1 = new ListNode(0);}if (l2 == NULL) {l2 = new ListNode(0);}                        temp = l1->val + l2->val;                        /*若上一位计算有进位，则当前结果加一*/            if (carry == 1) {                temp += 1;            }                        /*根据当前位结果记录进位*/            if (temp >= 10) {                temp -= 10;                carry = 1;            } else {                carry = 0;            }                        ListNode *currsum = new ListNode(temp);            if (head == NULL) {                head = currsum;                sum = head;            } else {                sum->next = currsum;                sum = sum->next;            }            l1 = l1->next;            l2 = l2->next;        }                /*若最高位计算结果有进位，则在表尾插入新节点1*/        if (carry == 1) {        sum->next = new ListNode(1);}        return head;    }};