[leetcode]25. Reverse Nodes in k-Group

最新推荐文章于 2025-05-21 19:43:17 发布

xiaocong1990

最新推荐文章于 2025-05-21 19:43:17 发布

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本文提供了一种解决LeetCode上K组翻转链表问题的方法。该问题要求将链表每K个节点进行翻转，并返回修改后的链表。文章详细介绍了如何实现这一翻转操作，包括递归地处理剩余节点，以及如何在不改变节点值的情况下仅改变节点本身来完成任务。

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题目链接：https://leetcode.com/problems/reverse-nodes-in-k-group/description/

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

k is a positive integer and is less than or equal to the length of the linked 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

class Solution
{
public:

    ListNode* reverse(ListNode* first, ListNode* last)
    {
        ListNode* new_head = new ListNode(-1);
        ListNode*p=first;

        while ( p != last )
        {
            ListNode*t=p;
            p=p->next;
            t->next=new_head->next;
            new_head->next=t;
        }

        return new_head->next;
    }

    ListNode* reverseKGroup(ListNode* head, int k)
    {
        auto node=head;
        for (int i=0; i < k; ++i)
        {
            if ( ! node  )
                return head; // nothing to do list too sort
            node = node->next;
        }

        auto new_head = reverse( head, node);
        head->next = reverseKGroup( node, k);
        return new_head;
    }
};