1 - Add Two Numbers-优快云博客

本文链接：https://blog.youkuaiyun.com/superrobin/article/details/11364933

本文介绍如何利用C++实现两个反转顺序的链表代表的非负整数相加，并返回结果作为新的链表。代码示例清晰易懂。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(l1 == NULL || l2 == NULL)
            return NULL;
        
        int carry = 0;
        int sum = 0;

        ListNode *resHead = NULL;
        ListNode *node = NULL;

        while(l1 != NULL || l2 != NULL)
        {
            if(l1 != NULL && l2 != NULL)
            {
                sum = l1->val + l2->val + carry;
                l1 = l1->next;
                l2 = l2->next;
            }
            else if(l1 == NULL && l2 != NULL)
            {
                sum = l2->val + carry;
                l2 = l2->next;
            }
            else if(l2 == NULL && l1 != NULL)
            {
                sum = l1->val + carry;
                l1 = l1->next;
            }
            
            carry = 0;
            
            if(sum >= 10)
            {
                sum = sum%10;
                carry = 1;
            }
            
            ListNode *newNode = new ListNode(sum);
            
            if(resHead == NULL)
            {
                resHead = newNode;
                node = resHead;
            }
            else
            {
                node->next = newNode;
                node = node->next;
            }
        }
        
        if(carry == 1)
        {
            sum = carry;
            node->next = new ListNode(sum);
            carry = 0;
        }
        
        return resHead;
    }
};