leetcode 445. Add Two Numbers II

1.题目

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
两个非空的链表来表示两个非负整数，计算两个数的和，结果也用链表来表示。
You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:
Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7
7243+564=7807

2.分析

一般来说，两个数的加法需要从最低位开始计算，然而单向链表只有向后的指针。
两种方案
1）用栈或者向量来存储每个链表的数字，达到可以从后往前访问的目的。 需要额外空间。
2）从前往后计算，先按节点存储节点之和，然后计算进位。需要翻转结果链表。

3.代码

方案2

class Solution {public:    ListNode* addNode(int value, ListNode* tail) {        ListNode* tmp = new ListNode(value);        tmp->next = tail;        return tmp;    }    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        if (l1 == NULL)            return l2;        if (l2 == NULL)            return l1;        ListNode *list1 = l1, *list2 = l2, *ans = NULL;        int len1 = 0, len2 = 0;        while (list1) {            list1 = list1->next;            ++len1;        }        while (list2) {            list2 = list2->next;            ++len2;        }        list1 = l1;        list2 = l2;        while (len1 > len2) {            ans = addNode(list1->val, ans);            --len1;            list1 = list1->next;        }        while (len2 > len1) {            ans = addNode(list2->val, ans);            --len2;            list2 = list2->next;        }        while (list1&&list2) {            ans = addNode(list1->val + list2->val, ans);            list1 = list1->next;            list2 = list2->next;        }        list1 = ans;        ans = NULL;        int carry = 0;        while (list1) {            list1->val += carry;            carry = list1->val / 10;            ans = addNode(list1->val%10, ans);            list1 = list1->next;        }        if (carry)            ans = addNode(1, ans);        return ans;    }};