【LeetCode & 剑指offer刷题】链表题9：Add Two Numbers

【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）

Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse orderand 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.

---

C++

 

//加链表表示的两个数，高位在链表后面，低位在前面

/**

 * Definition for singly-linked list.

 * struct ListNode {

 *     int val;

 *     ListNode *next;

 *     ListNode(int x) : val(x), next(NULL) {}

 * };

 */

//紧凑

class Solution

{

public:

    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)

    {

        ListNode preHead(0);

        ListNode* p = &preHead; //头结点,用于保存首结点指针，以及初始化p

        int carry = 0; //进位

       

        while(l1 || l2 || carry) //循环

        {

            int sum = (l1? l1->val:0) + (l2? l2->val:0) + carry; //对应位相加，判断是否为空，若为空，相当于加0

            carry = sum/10; //保存进位

            p->next = new ListNode(sum%10); //创建新的结点，并赋值，构造函数完成val和next指针的赋值

            p = p->next; //指向下一个结点

  

            l1 = l1? l1->next:l1; //若为空，则仍为空，若不为空，指向下一个结点

            l2 = l2? l2->next:l2;

           

        }

        return preHead.next; //返回首结点指针

       

    }

};