Add Two Numbers问题及解法

问题描述：

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

以下是问题分析：

本题是将两个数按照逆序存放在两个链表里，然后求这两个数的和，同样以逆序的方式输出为链表。此题中可能遇到的陷阱==》变量的生存周期问题（ListNode的某一变量在addTwoNumbers函数内使用，出了该范围无效）。

以下是我按照普通套路些的代码

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2){		ListNode* res = l1,*temp;			int flag = 0;			int tempval;						while(l1 != NULL && l2 != NULL){				tempval = l1->val + l2->val + flag;				flag = tempval / 10;				tempval = tempval % 10;				l1->val = tempval;				temp = l1;				l1 = l1->next;				l2 = l2->next;			}			if(l1 == NULL){			    temp->next = l2;				l1 = l2;			}			while(l1 != NULL){				tempval = l1->val + flag;				flag = tempval / 10;				tempval = tempval % 10;				l1->val = tempval;				temp = l1;				l1 = l1->next;			}			if(flag){				ListNode * p = (ListNode *)malloc(sizeof(ListNode));				ListNode f(flag);				p->val = f.val;				p->next = f.next;				temp->next = p; 			}			return res;		}
欢迎小伙伴们一起探讨~