leetcode-2 Add Two Numbers

leetcode-2 Add Two Numbers

思路：
1. 定义两个游标迭代器，分别初始化为指向链表头部l1,l2；
2. 从头部开始扫描l1和l2，两个游标指示元素相加之和取个位与进位相加，得到的值作为当前位的结果保存，并根据实际情况进行进位操作，如果某个游标先结束，则该游标及其后继都被标记为NULL；
3. 判断最后一位是否有进位，如果有，把最后一个进位最为最高位保存下来；
4. 返回最终链表的真实头指针（代码中加入了一个空的头结点，是为了便于代码迭代步骤的统一性）；

源码如下：

/*** 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* cursor1 = l1;        ListNode* cursor2 = l2;        ListNode* cursor3 = new ListNode(0);        ListNode* l3 = cursor3;        ListNode* temp = NULL;        int val1=0,val2 = 0;        bool incre = false;        int lowAdd = 0;        while (cursor1 != NULL || cursor2 != NULL) {            val1 = cursor1!=NULL?cursor1->val : 0;            val2 = cursor2!=NULL?cursor2->val :0;            temp = new ListNode(0);            temp->val = val1 + val2 + lowAdd;            temp->val <10 && (incre = false);            temp->val >= 10 && (temp->val -= 10, incre = true);            cursor3->next = temp;            cursor3 = temp;            temp = NULL;            lowAdd = incre ? 1 : 0;            cursor1 = cursor1!=NULL? cursor1->next : NULL;            cursor2 = cursor2!=NULL? cursor2->next : NULL;        }        incre && (cursor3->next = new ListNode(lowAdd));        cursor3= NULL;        l3 = l3->next;        delete cursor3;        //reverseNodes(l1);        return l3;    }    void reverseNodes(ListNode* &l) {        ListNode* cursor = l;        ListNode* temp = NULL;        l = NULL;        while (cursor != NULL) {            temp = cursor->next;            cursor->next = l;            l = cursor;            cursor = temp;        }    }};