学步园

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

思路：先将两个链表分别遍历一遍，获得链表长度，然后进行相加，时间复杂度为O(N)。

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        ListNode* sum = new ListNode(INT_MAX);
        ListNode* root = sum;
        int m1 = 0;
        int m2 = 0;
        ListNode* ptr1 = l1;
        ListNode* ptr2 = l2;
        while(ptr1 != NULL)
        {
            ++m1;
            ptr1 = ptr1->next;
        }
        while(ptr2 != NULL)
        {
            ++m2;
            ptr2 = ptr2->next;
        }
        if (m1 < m2)
        {
            ptr1 = l1;
            l1 = l2;
            l2 = ptr1;
            m1 ^= m2;
            m2 ^= m1;
            m1 ^= m2;
        }
        ptr1 = l1;
        ptr2 = l2;
        int flag = 0,val;
        ListNode* pNode;
        while(ptr2 != NULL)
        {
            val = (ptr1->val + ptr2->val + flag ) % 10;
            flag = (ptr1->val + ptr2->val + flag) / 10;
            pNode = new ListNode(val);
            sum->next = pNode;
            sum = sum->next;
            ptr2 = ptr2->next;
            ptr1 = ptr1->next;
        }
        while(ptr1 != NULL)
        {
            val = (ptr1->val + flag) % 10;
            flag = (ptr1->val + flag) / 10;
            pNode = new ListNode(val);
            sum->next = pNode;
            sum = sum->next;
            ptr1 = ptr1->next;
        }
        if (flag != 0)
        {
            pNode = new ListNode(flag);
            sum->next = pNode;
            sum = sum->next;
        }
        return root->next;
    }
};