[LeetCode]Add Two Numbers

Q：

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

以前写过大整数求和，不过是用数组实现的，这里要求用链表（无头结点），思路是一样的，我就不再叙述了，贴上链接：点击打开链接

in addition，添上链表的创建便于测试，贴代码：

#include <iostream>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* p1 = l1;ListNode* p2 = l2;ListNode* l3 = NULL;ListNode* node =NULL;ListNode* r = l3;int flag = 0;int sum = 0;while (p1!=NULL&&p2!=NULL){sum = (p1->val + p2->val + flag) % 10;flag = (p1->val + p2->val + flag) / 10;node = new ListNode(sum);if (l3 == NULL){l3 = node;}else{r->next = node;}r = node;p1 = p1->next;p2 = p2->next;}while (p1 != NULL){sum = (p1->val + flag) % 10;flag = (p1->val + flag) / 10;node = new ListNode(sum);r->next = node;r = node;p1 = p1->next;}while (p2 != NULL){sum = (p2->val + flag) % 10;flag = (p2->val + flag) / 10;node = new ListNode(sum);r->next = node;r = node;p2 = p2->next;}if (flag == 1){node = new ListNode(1);r->next = node;r = node;}return l3;}ListNode* createList(){ListNode* head = NULL;ListNode* rear = head;;ListNode* p = NULL;int x;while (cin >> x){p = new ListNode(x);if (head == NULL)head = p;elserear->next = p;rear = p;}cin.clear();return head;}};int main() {Solution s;ListNode* l1 = s.createList();ListNode* l2 = s.createList();ListNode* l3 = s.addTwoNumbers(l1, l2);ListNode* l = l3;while (l != NULL){printf("%d ", l->val);l = l->next;}printf("\n");return 0;}