Reverse Nodes in K-Group_reversenodesinkgroup-优快云博客

本文链接：https://blog.youkuaiyun.com/a83610312/article/details/8555554

本文介绍了一种算法，该算法将给定链表中的节点每K个一组进行翻转，并返回修改后的链表。文章详细解释了如何在不改变节点值的情况下仅改变节点本身来实现这一操作，并提供了具体的代码示例。

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问题描述如下：

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

主要考察指针的操作，在纸上画个例子，细心点就好了：

下面是代码：注意pKth指的是第K个节点， pFirst 到 pKth 是需要反转的子链：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
 #define LN ListNode*

class Solution {
public:
    LN reverseKGroup(LN head ,int k)
{
	if (!head||!head->next||k<=1)
		return head;

	LN pFirst=head;
	LN pBefore=NULL;
	LN pKth=pFirst;
	while(1)
	{
		int cnt=k-1;
		while(cnt&&pKth)
		{
			pKth=pKth->next;
			cnt--;
		}
		if ( cnt>0 || pKth==NULL)
			break;
		reverse(pBefore,pFirst,pKth);
		if (pBefore==NULL)
			head=pFirst;
		pBefore=pKth;
		pFirst=pKth->next;
		pKth=pFirst;
	}
	return head;
}
void reverse(LN pBefore,LN& pFirst,LN& pKth)
{
	LN pN=pFirst;
	while(pN!=pKth)
	{
		LN pTmp=pN->next;
		pN->next=pKth->next;
		pKth->next=pN;
		pN=pTmp;
	}
	LN pTmp=pFirst;
	pFirst=pKth;
	pKth=pTmp;
        if (pBefore)
            pBefore->next=pFirst;
}

		
};