Add Two Numbers 题解

题目来源：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

解题思路：

递归版

    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2, int carry)
    {
        if(l1 == NULL && l2 == NULL && carry == 0) //only comes the carry,like 5, 5 = 10, or carry = 0
            return NULL;
        ListNode * result = new ListNode(carry);
        if(l1 != NULL)
            result->val += l1->val;
        if(l2 != NULL)
            result->val += l2->val;
        result->next = addTwoNumbers(l1 == NULL ? NULL : l1->next, l2 == NULL ? NULL : l2->next, result->val / 10);
        result->val %= 10;
        return result;
    }

    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)
    {
        if(l1 == NULL && l2 == NULL)
            return NULL;
        if(l1 == NULL)
            return l2;
        if(l2 == NULL)
            return l1;
        return addTwoNumbers(l1, l2, 0);
    }

迭代版

    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) 
    {
        if(l1 == NULL && l2 == NULL) return NULL;
        if(l1 == NULL || l2 == NULL) return l1 == NULL ? l2 : l1;
        ListNode dummy(-1);
        ListNode * pre = &dummy;
        while(l1 && l2)
        {
            pre->next = new ListNode(l1->val + l2->val);
            pre = pre->next;
            l1 = l1->next;
            l2 = l2->next;
        }
        while(l1){
            pre->next = new ListNode(l1->val);
            pre = pre->next;
            l1 = l1->next;    
        }
        while(l2){
            pre->next = new ListNode(l2->val);
            pre = pre->next;
            l2 = l2->next;    
        }
        pre = dummy.next;
        while(pre){
            int t = pre->val;
            if(t >= 10) {
                pre->val = t % 10; 
                if(pre->next)
                    pre->next->val += t / 10;
                else
                    {pre->next = new ListNode(t / 10); break;}
            }
            pre = pre->next;
        }
        return dummy.next;
    }