LeetCodeOJ——2.Add Two Numbers

Add Two Numbers

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

解题思路：主要考链表和链表合并，尾插法（一开始用了头插法，尾插法链表的的建立，一开始建立了一个无效的头结点，返回的时候注意将这个无效的头结点删去）。

代码如下：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        ListNode* res = new ListNode(0);        res->next = NULL;        ListNode* q = res;        int add = 0;        while (l1!=NULL && l2!=NULL){            ListNode *tmpNode=new ListNode((l1->val + l2->val + add) % 10);            tmpNode->next = NULL;            add = (l1->val + l2->val + add) / 10;            q->next = tmpNode;            q = q->next;            l1=l1->next;            l2=l2->next;        }        while (l1!=NULL){            ListNode *tmpNode = new ListNode((l1->val + add) % 10);            tmpNode->next = NULL;            add = (l1->val + add) / 10;            q->next = tmpNode;            q = q->next;            l1=l1->next;        }        while (l2!=NULL){            ListNode *tmpNode = new ListNode((l2->val + add) % 10);            tmpNode->next = NULL;            add = (l2->val + add) / 10;            q->next = tmpNode;            q = q->next;            l2=l2->next;        }        if(add!=0){            ListNode *tmpNode = new ListNode(add);            tmpNode->next = NULL;            q->next = tmpNode;            q = q->next;        }        ListNode *p = res;        q = res->next;        delete p;        return q;    }};