leetcode Add Two Numbers

Add Two Numbers

题目详情：

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

解题方法：

这道题难度程度为medium，主要考查对加法进位的理解和对指针的应用。设第一个链表为L1，第二个链表为L2，得到答案新链表为L3。首先初始化进位为0，L3的第一位为L1，L2的第一位加上进位在取余10，进位则变为L1，L2的第一位加上上一次的进位除以10。主要方法是这样，但是要注意L1和L2的长度可能不相等，以及最后一位的进位处理。
我分为三种情况处理：L1比L2长，L2比L1长以及L1，L2一样长，每种情况中都还要考虑，最后一位加上进位后与10的大小比较。

题目详细代码：

/** * 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* l3;        ListNode* temp1 = l1;        ListNode* temp2 = l2;        ListNode* temp3 = new ListNode((l1->val + l2->val)%10);        l3 = temp3;        int jinwei = 0;        while(temp1->next != NULL && temp2 ->next != NULL) {           jinwei = (jinwei + temp1->val + temp2->val)/10;           temp3->next = new ListNode((jinwei+temp1->next->val+temp2->next->val)%10);           temp1 = temp1->next;            temp2 = temp2->next;            temp3 = temp3->next;        }        if (temp1->next == NULL && temp2->next != NULL) {            jinwei = (jinwei + temp1->val + temp2->val)/10;            temp3->next = new ListNode((jinwei+temp2->next->val)%10);            temp3 = temp3->next;            temp2 = temp2->next;            while(temp2 ->next != NULL) {           jinwei = (jinwei + temp2->val)/10;           temp3->next = new ListNode((jinwei+temp2->next->val)%10);            temp2 = temp2->next;            temp3 = temp3->next;}            if (jinwei + temp2->val >= 10) {                temp3->next = new ListNode((jinwei + temp2->val) / 10);            }        }else if (temp1->next != NULL && temp2->next == NULL) {            jinwei = (jinwei + temp1->val + temp2->val)/10;            temp3->next = new ListNode((jinwei+temp1->next->val)%10);            temp3 = temp3->next;            temp1 = temp1->next;            while(temp1 ->next != NULL) {           jinwei = (jinwei + temp1->val)/10;           temp3->next = new ListNode((jinwei+temp1->next->val)%10);            temp1 = temp1->next;            temp3 = temp3->next;}            if (jinwei + temp1->val >= 10) {                temp3->next = new ListNode((jinwei + temp1->val) / 10);            }        } else {           if (jinwei + temp2->val + temp1->val>= 10) {                temp3->next = new ListNode((jinwei + temp1->val+temp2->val) / 10);            }        }        return l3;    }};