Add Two Numbers--LeetCode

Add Two Numbers

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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

solution：

/** * 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) {        if (l1 == NULL && l2 == NULL) return NULL;        else if (l1 == NULL) return l2;        else if (l2 == NULL) return l1;                ListNode total(0);        ListNode* now = &total;        int jinwei = 0;        ListNode* L1 = l1;        ListNode* L2 = l2;                while(L1 || L2 || jinwei)        {            int d1 = L1 ? L1->val : 0;            int d2 = L2 ? L2->val : 0;            int a = (d1 + d2 + jinwei) % 10;            jinwei = (d1 + d2 + jinwei) / 10;            if (L1) L1 = L1->next;            if (L2) L2 = L2->next;            ListNode* newnode = new ListNode(a);            now->next = newnode;            now = now->next;        }        return total.next;    }};

另外，我们可以用递归解决这个问题。但是在用递归的时候，要注意进位的处理，在得出的结果中要再加上一。

/** * 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) {        if (l1 == NULL && l2 == NULL) return NULL;        else if (l1 == NULL) return l2;        else if (l2 == NULL) return l1;                int num = l1->val + l2->val;        ListNode* total = new ListNode(num % 10);        total->next = addTwoNumbers(l1->next, l2->next);        //处理进位        if (num >= 10)        {            total->next = addTwoNumbers(total->next, new ListNode(num / 10));        }        return total;    }};