Add Two Numbers

• 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

  ---

• A这道题主要是考虑到输入的两个连白哦是否为空， 长度不一致的处理，最关键的还是进位的处理。C++的代码如下：

/**
 * 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) { 
    ListNode *head=NULL, *curNode = NULL;
    if(!l1)
        return l2;
    if(!l2)
        return l1;

    int bit = 0;

    while(l1 && l2)
    {
        ListNode *newNode = new ListNode(l1->val + l2->val + bit);
        newNode->next = NULL;

        l1 = l1->next;
        l2 = l2->next;

        if(newNode->val >= 10)
        {
            newNode->val -= 10;
            bit = 1;
        }
        else
            bit = 0;

        if(!head)
        {
            head = newNode;
            curNode = head;
        }
        else
        {
            curNode ->next = newNode;
            curNode = curNode ->next;
        }
    }

    while(l1)
    {
        l1->val += bit;
        if(l1->val>=10)
        {
            l1->val -= 10;
            bit = 1;
        }
        else
        {
            bit = 0;
        }

        ListNode *newNode = new ListNode(l1->val);
        newNode->next = NULL;
        curNode->next = newNode;
        curNode=curNode->next;
        l1 = l1->next;
    }

    while(l2)
    {
        l2->val += bit;
        if(l2->val >=10)
        {
            l2->val -= 10;
            bit = 1;
        }
        else
        {
            bit = 0;
        }

        ListNode *newNode = new ListNode(l2->val);
        newNode->next = NULL;
        curNode->next = newNode;
        curNode=curNode->next;
        l2 = l2->next;
    }

    if(bit)
    {
        ListNode *tailNode = new ListNode(1);
        curNode->next = tailNode;
        curNode = curNode->next;
    }

    return head; 
    }
};