leetcode-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

题意：用链表模拟高精度加法。给出两个链表，分别逆序存储了两个数，即表头是个位上的数，第二节点是十位上的数……对其求和，结果以同样形式表示

分析：比较简单的链表操作题，每次遍历每个节点，计算每位上的和，记录进位，然后加入到新链表即可

注意：

1、addNode子函数是更新指针本身的值，所以参数要传递指针的引用

2、若两个链表长度不等，遍历完公共部分后要遍历较长的链表

3、最后若还有进位，要把进位加入结果链表

代码：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    void addNode(ListNode *&head, ListNode *&tail, ListNode *&now)    {        if(!head)        {            head = tail = now;        }        else        {            tail->next = now;            tail = now;        }    }    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {        ListNode *head=NULL,*tail=NULL,*now;        ListNode *t1=l1,*t2=l2;        int carry = 0;        for(; t1&&t2; t1=t1->next,t2=t2->next)        {            int value = t1->val+t2->val+carry;            carry = value/10;            value %= 10;            now = new ListNode(value);            addNode(head,tail,now);        }        for(; t1; t1=t1->next)        {            int value = t1->val+carry;            carry = value/10;            value %= 10;            now = new ListNode(value);            addNode(head,tail,now);        }        for(; t2; t2=t2->next)        {            int value = t2->val+carry;            carry = value/10;            value %= 10;            now = new ListNode(value);            addNode(head,tail,now);        }        if(carry)        {            now = new ListNode(carry);            addNode(head,tail,now);        }        return head;    }};