【LeetCode】2.Add Two Numbers两个单链表相加

You are given two non-empty linked lists（非空链表） representing two non-negative integers（非负整数）. 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
Input: (9->9) + (1)
Output: 0->0->1

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该题要求考虑进位。测试用例：链表长度不相等时；最后一次相加有进位时等。
注意： 如何new一个新的链表节点；设置p指针时不能少了*号；->的使用，而不是用 . 。

/** * 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;        ListNode* p2 = l2;        ListNode* l3 = new ListNode(0);        ListNode* p3 = l3;        int carry = 0;//进位        while(p1 != NULL || p2 != NULL){    //只要还有一个链表不为空，就计算sum；而且不要去判断p1->next是不是为空！            int sum = 0;        //还有种不用谢carry的写法：while外面写int sum=0；里面写sum /= 10;            if(p1 != NULL){                sum += p1->val;                p1 = p1->next;            }            if(p2 != NULL){                sum += p2->val;                p2 = p2->next;            }            sum += carry;            carry = sum/10;            p3->next = new ListNode(sum % 10);            p3 = p3->next;        }        if(carry == 1){//最后一个值还要进位            p3->next = new ListNode(1);        }        return l3->next;        //return l3;    //错误结果为[0,7,0]    }};

class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        ListNode* l3 = new ListNode(0);        ListNode* p = l3;        int carry = 0;  //进位        while(l1 || l2 || carry){    //l1、l2、进位均不为空时            int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + carry;//如果l1不空取当前val值，否则赋值为0            carry = sum/10;            p->next = new ListNode(sum%10);            p = p->next;            l1 = l1 ? l1->next : l1;    //如果l1不空往下移，否则不移动            l2 = l2 ? l2->next : l2;        }        return l3->next;    }};