2. 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

Subscribe to see which companies asked this question.

题解：这道题目的本身设计比较人性化，因为他是将两个数字存到两个链表中进行相加，因为是倒序存储，所以，我们正好可以从链表的头结点处开始相加，初始设置进位为0，从前往后相加，满10进1，直到两个链表同时为空和进位为0为止

class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        ListNode head(0);        ListNode *pt = &head;        int carry = 0;        int sum;        while(l1 != NULL|| l2 != NULL || carry != 0){            sum = (l1 != NULL?l1->val:0)+(l2 != NULL?l2->val:0) + carry;            pt->next = new ListNode(sum % 10);            pt = pt->next;            carry = sum / 10;            l1 = l1 !=NULL?l1->next:NULL;            l2 = l2 !=NULL?l2->next:NULL;        }                return head.next;    }};