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最新推荐文章于 2026-01-05 22:36:07 发布

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1.题目

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 contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example 1:

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

Example 2:

Input: l1 = [0], l2 = [0]
Output: [0]

Example 3:

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]

Constraints:

The number of nodes in each linked list is in the range [1, 100].
0 <= Node.val <= 9
It is guaranteed that the list represents a number that does not have leading zeros.

2.思路

1.要考虑有进位的情况比如十进制 “/10” 以及“%10”

2. 创建一个新的链表来接受结果，最后返回头节点即可

3.代码

#include <iostream>
#include <memory>


using namespace std; 
struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr) {}
};

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        // 要设置加法的标志位 node的值在0-9 之间 加的值/10 为 标志位满  
        //特殊情况就是当到了链表的最后一个节点了 加法标志位仍然为1 要像结果最后面加个1  
        int flag = 0 ;
        ListNode* dummy = new ListNode(0); //创建虚拟的头节点
        ListNode*  cur = dummy; //使用指针指向


        while (l1 != nullptr || l2 != nullptr)
        {
            int sum = flag;

            if (l1 != nullptr)
            {
                sum += l1->val ;
                l1 = l1->next;
            }

            if (l2 != nullptr)
            {
                sum += l2->val ;
                l2 = l2->next;
            }

          
            flag = (sum / 10);
            cur->next = new ListNode(sum%10);
            cur = cur->next; 

        }

        if (flag > 1)
        {
            cur->next = new ListNode(flag);
        }
        return dummy->next; //返回真正的头节点
        
    }
};

int main() {
    // 创建第一条链表：2 -> 4 -> 3
    ListNode* l1 = new ListNode(2);
    l1->next = new ListNode(4);
    l1->next->next = new ListNode(3);

    // 创建第二条链表：5 -> 6 -> 4
    ListNode* l2 = new ListNode(5);
    l2->next = new ListNode(6);
    l2->next->next = new ListNode(4);

    Solution s1;


    // 输出结果链表
    ListNode* result = s1.addTwoNumbers(l1, l2);
    while (result != nullptr) {
           
        std::cout << result->val << " ";
         result = result->next;
    }
    std::cout << std::endl;

    return 0;
}