LeetCode OJ 2.Add Two Numbers

最新推荐文章于 2019-03-08 21:35:11 发布

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最新推荐文章于 2019-03-08 21:35:11 发布

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分类专栏： LeetCode OJ

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本文介绍了一种解决两链表代表的非负整数相加问题的方法。通过遍历链表，将数值转换为整数求和后再逆序转换回链表形式返回。此算法涉及链表操作及数值计算。

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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

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就是遍历，求值，再恢复

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
classSolution{
public:
                ListNode*addTwoNumbers(ListNode*l1, ListNode*l2) {
                                unsignedlonglongsum3= 0;
                                intcount1= 0;
                                intcount2= 0;
                                if(l1==NULL||l2==NULL)
                                {
                                                returnNULL;
                                }
                                while(l1!=NULL)
                                {
                                                sum3+=pow(10,count1) * (l1->val);
                                                count1++;
                                                l1=l1->next;
                                }
                                while(l2!=NULL)
                                {
                                                sum3+=pow(10,count2) * (l2->val);
                                                count2++;
                                                l2=l2->next;
                                }
                                unsignedlongtmp_digit=sum3- (sum3/ 10) * 10;
                                ListNode*p=newListNode((int)tmp_digit);
                                ListNode*start=p;
                                sum3=sum3/ 10;

                                while(sum3> 0) {
                                                tmp_digit=sum3- (sum3/ 10) * 10;
                                                p->next=newListNode((int)tmp_digit);
                                                p=p->next;
                                                sum3=sum3/ 10;
                                }
                                returnstart;
                }
};