[Leetcode] reverse nodes in K-Group

最新推荐文章于 2025-05-21 19:43:17 发布

原创最新推荐文章于 2025-05-21 19:43:17 发布 · 462 阅读

0 ·

CC 4.0 BY-SA版权

Leetcode 专栏收录该内容

67 篇文章

订阅专栏

本文详细介绍了如何通过引入两个指针 beforeK 和 afterK，利用传统反转链表的方法来解决给定链表中反转k个节点的问题。通过实例演示了如何在不改变节点值的情况下，实现对链表特定长度段的反转，并讨论了当链表长度不是k的倍数时的特殊情况处理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For example,
Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

这道题一开始怎么做也做不对。。。

后来换了种策略，设计两个指针，分别为 beforeK 和 afterK，再加上 reverse nodes 的传统做法就可以了。

这次做一次搞定。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *reverseKGroup(ListNode *head, int k) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        if(head==NULL) return NULL;
        ListNode *vHead = new ListNode(0);
        if(k==1) return head;
        
        vHead->next = head;
        ListNode *beforeK = vHead;
        ListNode *afterK = head;
        int count = 0;
        
        
        while(true)
        {
            while(count<k&&afterK!=NULL)
            {                
                afterK = afterK->next;
                count++;
            }//whether there is enough nodes
            
            if(count!=k)
            {
                break;
            }
            
            ListNode* current = beforeK->next;
            ListNode* post = NULL;
            ListNode* pre = afterK;
            
            ListNode* newTail = current;
            
            while(current!=afterK)
            {
                post = current->next;
                current->next = pre;
                pre = current;
                current = post;
            }
            
            beforeK->next = pre;
            beforeK = newTail;
            count = 0;
        }
        
        return vHead->next;
    }
};