[LrrtCode] Add Two Numbers

题目

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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

分析

真的想不到我打这道题打了半个小时。。。。其实是个很简单的东西呀，就是链表相加，从左到右加起来就可以了，遇到满10的进一位。一开始我是直接转化成整数然后取回文数，然后相加，再生成链表，然后发现这样做会超出long long的表示范围。于是就改成将链表数据转移到两个数组当中，然后数组逐项相加，冯十进一就ok啦。注意结果的位数是不是多了一位。比如两个二维数相加结果可能是三位数。简单的问题，毫无算法可言。

代码

/**
 * 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* p1 = l1;
        ListNode* p2 = l2;
        int t[100] = {0};
        int t2[100] = {0};
        int size = 0;
        while(p1!=NULL)
        {
            t[size++] = p1->val;
            p1 = p1->next;
        }
        int s=0;
        while(p2!=NULL)
        {
            t2[s++] = p2->val;
            p2 = p2->next;
        }
        if(s>size) size=s;
        for(int i=0;i<size;i++)
        {
            t[i] = t[i] + t2[i];
            if(t[i]>=10)
            {
                t[i]%=10;
                t[i+1]+=1;
                if(i==size-1) size++;
            }
        }
        
        ListNode* p = NULL;
        ListNode* head = NULL;
        int n = 0;
        while(n<size)
        {
            if(n==0)
            {
                head = new ListNode(t[n]);
                p=head;
            }
            else
                p->next = new ListNode(t[n]);
            if(n!=0)
                p = p->next;
            ++n;
        }
        return head;
        
    }
};