2. Add Two Numbers

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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

/**   定义链表类数据结构 * 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* result = new ListNode(0); //通过new来申请动态内存,使用ListNode的构造函数初始化一个对象,表达式返回指向这个对象的指针        ListNode *r,*tmp;        int count = 0;        r = result;  //最后返回result,故result指针是不动的,将其赋值给r        while(l1&&l2) //非空        {            int isum = count + l1->val + l2->val;            count = isum/10;            tmp = new ListNode(isum%10);             r->next = tmp;//将r->next = tmp ,注意,结果链表的头节点不用来存值,而是从它的next开始存放,返回时返回result->next            r = r->next;            l1 = l1->next;            l2 = l2->next;        }        if(l1 == NULL&&l2!=NULL) //l1空,l2非空        {            r->next = l2; //将l2赋值给r->next            if(count == 0)   //如果上面的while循环结束时,count=0,即没有进位,则返回            {                return result->next;  //比如:l1: 2 3 / l2 1,3,5              }            else  //当有进位时,即count !=0,此时要考虑进位            {     // 比如: 2 3 /  1 7 5 / 3 0 6                while(r->next&&count!=0) //此处用r->next 先判断r的next是否为空                {                    r = r->next; //不为空,则赋值给r                    int isum = r->val + count;                    r->val = isum%10;                    count = isum/10;                }                if(r->next == NULL && count!=0) //当r为空时,但是,此时上面的while返回的count!=0,则还想考虑                {   // 例如: 1 / 9 9 /0 0 1                    tmp = new ListNode(count);                    r->next = tmp;                    return result->next;                }                else return result->next;            }        }        else if(l1 !=NULL&&l2==NULL)        {           r->next = l1;           if(count == 0)            {                return result->next;            }            else            {                while(r->next&&count!=0)                {                    r = r->next;                    int isum = r->val + count;                    r->val = isum%10;                    count = isum/10;                }                if(r->next == NULL && count!=0)                {                    tmp = new ListNode(count);                    r->next = tmp;                    return result->next;                }                else return result->next;            }        }        else        {               if(count == 0)                return result->next;            else            {   // count !=0,需要再次申请节点 2 3 4 / 1 3 6 / 3 6 0 1                tmp = new ListNode(count);                r->next = tmp;                return result->next;            }        }    }};

总结:
此题为链表的操作:
1.因为定义类型指针如ListNode *tmp ,指针变量占4字节(32位机),再申请动态内存对象时,注意使用tmp = new ListNode() ,将tmp指针指向动态分配的对象,然后再进行后续操作,最后delete它
2.while中一般多用while(p->next)来判断,如果进入while循环,再将p = p->next;
3.此题中,返回result时,不是直接返回result,而是返回它的next;因为这样处理使编写程序的过程中简单点,需要借鉴
4.此题中,当l1或l2为空时,要考虑到count!=0情况;

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