LeetCode OJ:Add Two Numbers

Description

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

Code

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */ //此题考虑要全面：l1和l2两个链表不一定等长；最后两个数相加之后是否为两位数。为减少内存分配，直接把计算结果存放在l1或者l2中即可。class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        ListNode* pt,*s;  //pt用来循环，s总是指向pt的后一个位置        pt = l1;        int m,n=0;  //m,n分别记录两个数相加之后的个位和十位        while(pt&&l2){            m = pt->val + l2->val + n;            n = m /10;            m = m %10;            pt->val = m;            l2 = l2->next;            s= pt;            pt = pt->next;        }        if(pt){ //l1比l2长             while(pt){                m = pt->val + n;                n = m / 10;                m=m%10;                pt->val = m;                s= pt;                pt = pt->next;             }        }else if(l2){ //l2比l1长             while(l2){                 m = l2->val + n;                n = m / 10;                m=m%10;                l2->val = m;                s->next= l2;                s = l2;                l2 = l2->next;             }         }         if(n==1){ //最后两个数相加为两位数，要另外分配一个存储块。            ListNode* p=new ListNode(0);            p->val = n;            p->next = NULL;            s->next = p;        }        return l1;    }};