Add Two Numbers我见过的最简洁的C++实现之我的第一篇博客

 ListNode preHead(0), *p = &preHead;    int extra = 0;    while (l1 || l2 || extra) {        int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + extra;        extra = sum / 10;        p->next = new ListNode(sum % 10);        p = p->next;        l1 = l1 ? l1->next : l1;        l2 = l2 ? l2->next : l2;    }    return preHead.next;

leetCode算法题目：Add Two Numbers   这是我见过的最简洁的C++实现了，作为学习笔记整理出来，方便日后查看

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

我的第一版：

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {ListNode *l3 = new ListNode();ListNode *l5;l5 = l3;int temp = 0;if (l1 != NULL){l3->val = ((l1->val) + (l2->val)) % 10 + temp;   //将节点相加temp = ((l1->val) + (l2->val)) / 10;l3->next = (ListNode*)malloc(sizeof(struct ListNode));l3 = l3->next;l1 = l1->next;l2 = l2->next;}return l5;}

提示内存不对齐，为失败版本