445. Add Two Numbers II。

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.

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

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这个题跟Add Two Numbers是一样的，但是这个是需要从链表的最后面开始进行相加。我的思路就是使用栈的特性，遍历两个链表，然后分别将两个链表中的内容存放到两个栈中，然后再根据栈中的数据分别出栈，这样每次出栈的元素的顺序就是原先链表从后往前的顺序。再使用跟之前那个一样的算数体系，使用一个变量保存两个数的总和，一个变量保存进位的数值。不断的创建新的节点，这里需要使用头插法来连接链表。

#include <iostream>#include <stack>using namespace std;struct ListNode {    int val;    ListNode *next;    ListNode(int x) : val(x), next(NULL) {}};class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        ListNode* result = new ListNode(0);        stack<int> nodes1;        stack<int> nodes2;        int carry = 0;        int sum;        while(l1 || l2) {            if(l1) {                nodes1.push(l1->val);                l1 = l1->next;            }            if(l2) {                nodes2.push(l2->val);                l2 = l2->next;            }        }        int n1,n2;        while(!nodes1.empty() || !nodes2.empty()) {//因为两个栈的元素一样多            if(nodes1.empty()) {                n1 = 0;            } else {                n1 = nodes1.top();                nodes1.pop();            }            if(nodes2.empty()) {                n2 = 0;            } else {                n2 = nodes2.top();                nodes2.pop();            }            //cout << n1 << "," << n2 << endl;            sum = n1 + n2 + carry;            //cout << "sum:" << sum << endl;            carry = sum / 10;//进的位数            result->val = sum % 10;//最后面一个节点赋值            ListNode* node = new ListNode(0);            node->next = result;            result = node;        }        if(carry) {            result->val = carry;            return result;        } else {            return result->next;        }    }};int main() {    Solution s;    ListNode node1(7);    ListNode node2(2);    ListNode node3(4);    ListNode node4(3);    ListNode node5(5);    ListNode node6(6);    ListNode node7(4);    node1.next = &node2;    node2.next = &node3;    node3.next = &node4;    node5.next = &node6;    node6.next = &node7;    ListNode* p = s.addTwoNumbers(&node1,&node5);    while(p) {        cout << p->val << endl;        p = p->next;    }}

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运行结果可能有误差。