Add Two Numbers算法

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

这道题，一开始我是想投机取巧的，想先提取出342和564，然后相加等于807，再逐个放进ListNode里面的。

经过一系列的DeBug之后，我觉得已经ok了。然而......

我也很难受啊，可我能怎么办嘛。怎么办？重打呗！

正题部分来了

既然这种投机取巧不行，那就老老实实地逐位相加吧

/**
 * 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;
        ListNode* temp ;
        int a = 0, b = 0,carry = 0;
        while(l1) {
            a++;
            l1 = l1->next;
        }
        while(l2) {
            b++;
            l2 = l2->next;
        }
        if(a >= b){
            temp = p1;
            while(p1){
                if(p2){
                    p1->val = p1->val + p2->val + carry;
                    carry = 0;
                    if(p1->val >= 10) {
                        p1->val -= 10;
                        carry = 1;
                    }
                    p2 = p2->next;
                }
                else {
                    p1->val = p1->val + carry;
                    carry = 0;
                    if(p1->val >= 10) {
                        p1->val -= 10;
                        carry = 1;
                    }
                }
                if(p1->next == NULL && carry == 1) {
                    p1->next = new ListNode(1);
                    break;
                }
                p1 = p1->next;
            }
        }
        else{
            temp = p2;
            while(p2){
                if(p1){
                    p2->val = p2->val + p1->val + carry;
                    carry = 0;
                    if(p2->val >= 10) {
                        p2->val -= 10;
                        carry = 1;
                    }
                    p1 = p1->next;
                }
                else {
                    p2->val = p2->val + carry;
                    carry = 0;
                    if(p2->val >= 10) {
                        p2->val -= 10;
                        carry = 1;
                    }
                }
                if(p2->next == NULL && carry == 1) {
                    p2->next = new ListNode(1);
                    break;
                }
                p2 = p2->next;
            }
        }
        return temp;
    }
};

然后呢，大概就是这样，要注意的地方就是进位那里了，这里我用carry来表示进位。

还有，就是空指针，比如说，3-4-5-6 加上1-2-3，就不能一味地使用 p1->val + p2->val + carry，当第4位相加时，p2的第4位已然是空的，就不能再调用p2了，这样会出现空指针的错误。