Add Two Numbers

链表加法：两数相加并返回结果链表

最新推荐文章于 2025-09-04 16:11:29 发布

转载最新推荐文章于 2025-09-04 16:11:29 发布 · 49 阅读

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原文链接：http://www.cnblogs.com/pengyu2003/p/3581887.html

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#数据结构与算法

本文介绍了一种通过给定两个表示非负整数的链表来相加的方法，结果同样以链表形式返回。详细步骤包括初始化进位标志、遍历链表进行加法操作，并考虑链表长度不一致的情况。

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

简单链表题

/**
 * 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) {
        int flag = 0;
        if(l1 == NULL && l2 == NULL)return NULL;
 
        ListNode * head = new ListNode(0);
        ListNode * tmp =head;
        
        while(l1!=NULL && l2!=NULL)
        {
            int temp = l1->val + l2->val + flag;
            flag = temp/10;
            ListNode *node = new ListNode(temp%10);
            tmp->next = node;
            tmp = tmp -> next;
            l1 = l1->next;
            l2 = l2->next;
        }
        while(l1!=NULL)
        {
            int temp = l1->val + flag;
            flag = temp/10;
            ListNode *node = new ListNode(temp%10);
            tmp->next = node;
            tmp = tmp -> next;
            l1 = l1->next;
        }
        while(l2!=NULL)
        {
            int temp = l2->val + flag;
            flag = temp/10;
            ListNode *node = new ListNode(temp%10);
            tmp->next = node;
            tmp = tmp -> next;
            l2 = l2->next;
        }
        if(flag == 1)
        {
            ListNode *node = new ListNode(1);
            tmp->next = node;
        }
        head =head->next;
        return head;
        
    }
};

转载于:https://www.cnblogs.com/pengyu2003/p/3581887.html