LeetCode 题解（26）: Add Two Numbers

最新推荐文章于 2025-08-16 00:47:01 发布

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#algorithm #leetcode #Add Two Numbers

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本文介绍了如何通过链表解决逆序整数相加问题，详细解释了算法实现过程，包括处理不同长度链表、进位逻辑及最终返回正确结果的步骤。

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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

题解：

虽然简单，但是要小心陷阱。如果没说是非负数，而且是用string表示，还要考虑正负号的问题，如果使用2进制表示，也要考虑正负号的问题。

还要考虑两个链表不一样长的问题。以及最后进位产生新节点的问题。

/**
 * 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* first = l1;
        ListNode* second = l2;
        
        if(!l1 && !l2)
            return NULL;
        else if(!l1)
            return l2;
        else if(!l2)
            return l1;
            
        int carry = (l1->val + l2->val) >= 10 ? 1 : 0;
        ListNode* l3 = new ListNode((l1->val + l2->val) % 10);
        ListNode* pre = l3;
        first = first->next;
        second = second->next;
        while(first && second)
        {
            ListNode* newNode = new ListNode((first->val + second->val + carry) % 10);
            carry = (first->val + second->val + carry) >= 10 ? 1 : 0;
            pre->next = newNode;
            pre = newNode;
            first = first->next;
            second = second->next;
        }
        
        if(!first)
        {
            while(second)
            {
                ListNode* newNode = new ListNode((second->val + carry) % 10);
                carry = (second->val + carry) >= 10 ? 1 : 0;
                pre->next = newNode;
                pre = newNode;
                second = second->next;
            }
            if(carry)
            {
                ListNode* newNode = new ListNode(1);
                pre->next = newNode;
            }
        }
        else if(!second)
        {
            while(first)
            {
                ListNode* newNode = new ListNode((first->val + carry) % 10);
                carry = (first->val + carry) >= 10 ? 1 : 0;
                pre->next = newNode;
                pre = newNode;
                first = first->next;
            }
            if(carry)
            {
                ListNode* newNode = new ListNode(1);
                pre->next = newNode;
            }            
        }
        
        return l3;
    }
};