leetcode 445. Add Two Numbers II

最新推荐文章于 2025-08-22 19:21:44 发布

原创最新推荐文章于 2025-08-22 19:21:44 发布 · 364 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #445 #add #two #numbers

Leetcode 专栏收录该内容

175 篇文章

订阅专栏

本文介绍了一种使用栈来辅助实现两个链表数值相加的方法。通过将链表节点值压入栈中，再逐个弹出进行相加，并处理进位逻辑，最终构建新的链表返回结果。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

用栈先存着两个node的东西，再pop出来相加。

转载请注明： http://blog.youkuaiyun.com/c602273091/article/details/54317115

/**
 * 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 cf = 0;
        stack<int> q_1;
        stack<int> q_2;


        while(l1 != NULL || l2 != NULL)
        {
            if(l1 != NULL)
            {
                q_1.push(l1->val);
                l1 = l1->next;
            }    
            if(l2 != NULL)    
            {
                q_2.push(l2->val);
                l2 = l2->next;
            }    
        }

        ListNode *rel = NULL;

        while(!q_1.empty() && !q_2.empty())
        {
            int a1 = q_1.top(), a2 = q_2.top();
            cout << a1 << "  " << a2 << endl;
            int sum_tmp = q_1.top() + q_2.top() + cf;
            q_1.pop();
            q_2.pop();
            cf = sum_tmp/10;
            sum_tmp = sum_tmp%10;
            ListNode *tmp = new ListNode(sum_tmp);
            tmp->next = rel;
            rel = tmp;
        }

        while(!q_1.empty())
        {
            int tmp_q_1 = q_1.top() + cf;
            q_1.pop();
            cf = tmp_q_1/10;
            tmp_q_1 = tmp_q_1 % 10;
            ListNode *tmp2 = new ListNode(tmp_q_1);
            tmp2->next = rel;
            rel = tmp2;            
        }

        while(!q_2.empty())
        {
            int tmp3 = q_2.top() + cf;
            q_2.pop();
            cf = tmp3/10;
            tmp3 = tmp3 % 10;
            ListNode *tmp2 = new ListNode(tmp3);
            tmp2->next = rel;
            rel = tmp2;             
        }

        if(cf)
        {
            ListNode *tmp = new ListNode(cf);
            tmp->next = rel;
            rel = tmp;
        }

        /*
        while(rel != NULL)
        {
            cout << rel->val << endl;
            rel = rel->next;
        }*/

        return rel;
    }
};