【Add Two Numbers】

题目：

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

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 dummy(-);
        ListNode *p = &dummy,*p1 = l1,*p2 = l2;
        int carry = ;
        while(p1!=NULL || p2!=NULL){
            const int v1 = p1==NULL?:p1->val, v2 = p2==NULL?:p2->val, v = (v1+v2+carry)%;
            carry = (v1+v2+carry)/;
            p->next = new ListNode(v);
            p = p->next;
            p1 = p1==NULL ? NULL : p1->next;
            p2 = p2==NULL ? NULL : p2->next;
        }
        if ( carry >  ) p->next = new ListNode(carry);
        return dummy.next;
    }
};

Tips：

核心在于判断while停止条件：直到l1和l2都走完了才退出；如果l1或者l2先走完了，就当该位是0。

上面这种思路的好处是可以简化代码。

=========================================

第二次过这道题，已经对思路比较熟悉了；指针定义操作什么的，稍微想了一下；代码还是一次AC。

/**
 * 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) {
        int carry = ;
        ListNode* pre = new ListNode();
        ListNode* head = pre;
        while ( l1 || l2 )
        {
            int digit1 = l1 ? l1->val : ;
            int digit2 = l2 ? l2->val : ;
            int curr_val = (digit1 + digit2 + carry)%;
            carry = (digit1 + digit2 + carry)/;
            if ( l1 ) l1 = l1->next;
            if ( l2 ) l2 = l2->next;
            pre->next = new ListNode(curr_val);
            pre = pre->next;
        }
        if ( carry> ) pre->next = new ListNode(carry);
        return head->next;
    }
};