Leetcode——2. Add Two Numbers

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) 
    {
        std::vector<int> num1, num2;   //存储数字1,2
        int num_count1(0), num_count2(0);   //数字数组1和2中数字的个数
        
        //获得链表1对应的数字
        ListNode* temp_node = l1;
        while(NULL != temp_node)
        {
            num1.push_back(temp_node->val);
            temp_node = temp_node->next;
        }
        num_count1 = num1.size();
        
        //获得链表2对应的数字
        temp_node = l2;
        while(NULL != temp_node)
        {
            num2.push_back(temp_node->val);
            temp_node = temp_node->next;
        }
        num_count2 = num2.size();
        
        //两个数字数组相加
        if(num_count1 > num_count2)
        {
            for(int i=num_count2; i<num_count1; i++)
            {
                num2.push_back(0);
            }
        }
        else if(num_count1 < num_count2)
        {
            for(int i=num_count1; i<num_count2; i++)
            {
                num1.push_back(0);
            }
        }
        
        //两个数字相加且分割
        std::vector<int> num3;   //两个数字相加得到的结果数组
        int num_count = num1.size();
        bool flag = false;  //数字相加进位标志
        for(int i=0; i<num_count; i++)
        {
            int temp = num1[i] + num2[i];
            if(flag)
            {
                temp++;
                flag = false;
            }
            if(temp > 9)
            {
                temp -= 10;
                flag = true;
            }
            num3.push_back(temp);
        }
        if(flag)
            num3.push_back(1);
        
        
        ListNode* return_node = new ListNode(0);
        temp_node = return_node;
        /*
        if(num_count >= 10)
        {
            num3.push_back(data1+data2);
            num3.push_back(data1);
            num3.push_back(data2);
            for(int i=0; i<num2.size(); i++)
                num3.push_back(num2[i]);
        }
        */
        num_count = num3.size();
		if(num_count > 0)
		{
			return_node->val = num3[0];
			if(num_count > 1)
			{
				for(int i=1; i<num_count; i++)
				{
					ListNode* temp = new ListNode(num3[i]);
					return_node->next = temp;
					return_node = return_node->next;
				}
			}
		}
		return temp_node;
    }
};