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

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/

很简单的基于链表的加法，这种算法常用于大整数的加法

class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        ListNode *temp1,*temp2,*result,*head,*pnext;        temp1 = l1;        temp2 = l2;        pnext=result = head = NULL;        bool carry = false;        if(l1 == NULL || l2 == NULL)            return head;        while(temp1 != NULL && temp2 != NULL)        {            pnext = new ListNode(0);            int sum = temp1->val+temp2->val;            if(carry == true)            {                sum += 1;                carry =false;            }            if(sum >= 10)            {                pnext->val = sum%10;                carry =true;            }            else            {                pnext->val = sum;                carry =false;            }            if(result == NULL)            {                head = result = pnext;            }            else            {                result->next = pnext;                result = pnext;            }            temp1 = temp1->next;            temp2 = temp2->next;        }       while(temp1 != NULL)        {            pnext = new ListNode(0);            if(carry == true)            {                pnext->val = (temp1->val+1)%10;                if(temp1->val+1 >= 10)                    carry =true;                else                    carry = false;            }            else            {                pnext->val = temp1->val;            }            temp1 = temp1->next;            result->next = pnext;            result = pnext;        }        while(temp2 != NULL)        {            pnext = new ListNode(0);            if(carry == true)            {                pnext->val = (temp2->val+1)%10;                if(temp2->val+1 >= 10)                    carry =true;                else                    carry = false;            }            else            {                pnext->val = temp2->val;            }            temp2 = temp2->next;            result->next = pnext;            result = pnext;        }        if(carry == true)        {            pnext = new ListNode(1);            result->next = pnext;        }        return head;    }};

改进方法将两个链表分别转换为对应的长整数num1,num2(虽然长整数很大，但仍然存在溢出风险),然后求出两个长整数的和sum，然后再将sum转换为对应的链表

class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        long num1=0;        long num2=0;        long mult = 1;        long sum=0;        ListNode *temp1,*temp2,*result,*head,*pnext;        temp1 = l1;        temp2 = l2;        head = result =pnext=NULL;        while(temp1!=NULL)        {            num1 += temp1->val*mult;            mult *= 10;            temp1 = temp1->next;        }        mult = 1;        while(temp2!=NULL)        {            num2 += temp2->val*mult;            mult *= 10;            temp2 = temp2->next;        }        sum=num1+num2;        if(sum == 0)        {            pnext = new ListNode(0);            head =result = pnext;        }        else        {          while(sum != 0 )          {                int val = sum%10;                sum =sum/10;                pnext = new ListNode(val);                if(head == NULL)                {                    head = result =pnext ;                }                else                {                    result->next = pnext;                    result = pnext;                }        }        }        return head;    }};