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


此题着实不难,然考虑甚多. 边界情况种种, 不一而足, 耗时亦两小时, 实感愧怍.

委屈

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
  /*  ListNode *reverseList(ListNode *head)
    {
        if(!head||!head->next)
            return head;
        ListNode *p,*q,*r;
        p = NULL;
        q = head;
        r = head->next;
        while(r)
        {
            q->next = p;
            p = q;
            q = r;
            r = r->next;
        }
        return q;
    }
    */
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) 
    {
        ListNode *p1, *p2;
        //p1 = reverseList(l1);
        //p2 = reverseList(l2);
        p1 = l1;
        p2 = l2;
        if(!p1) return p2;
        if(!p2) return p1;
        if(!p1->next&&!p2->next)
        {
            if(p1->val+p2->val<10)
            {
                p1->val += p2->val;
                return p1;
            }
            else
            {
                p1->val = (p1->val+p2->val)%10;
                p2->val = 1;
                p1->next = p2;
                return p1;
            }
        }
        int carry = 0;
        ListNode *p = NULL;
        ListNode *newHead = NULL;
        ListNode *pNext = NULL;
        while(p1&&p2)
        {
           pNext = new ListNode(0);
           if(!p)
            {
                p = pNext;
                newHead = p;
            }
            else
            {
                p->next  = pNext;
                p = pNext;
            }
               
            pNext->val = (p1->val+p2->val+carry)%10;
            carry  = (p1->val+p2->val+carry)/10;
            if(!p1->next&&!p2->next) //{5},{5}
            {
                if(carry == 1)
                {
                    pNext = new ListNode(1);
                    p->next  = pNext;
                    return newHead;
                }
            }
                p1 = p1->next;
                p2 = p2->next;
            }

        if(!p1)
        {
            while(p2)
            {
                pNext = new ListNode(0);
                pNext->val = (p2->val+carry)%10;
                carry  = (p2->val+carry)/10;
                p2 = p2->next;
                p->next  = pNext;
                p = pNext;
            }
            if(carry == 1)
            {
                 pNext = new ListNode(1);
                 p->next = pNext;
            }
        }
        if(!p2)
        {
            while(p1)
            {
                pNext = new ListNode(0);
                pNext->val = (p1->val+carry)%10;
                carry  = (p1->val+carry)/10;
                p1 = p1->next;
                p->next  = pNext;
                p = pNext;
            }
             if(carry == 1)
            {
                 pNext = new ListNode(1);
                 p->next = pNext;
            }
        }
        return newHead;
        
    }
};


posted @ 2013-12-13 06:40  海滨银枪小霸王  阅读(125)  评论(0)    收藏  举报