Leetcode217: Reverse Nodes in k-Group-优快云博客

本文介绍了一种算法，用于每K个一组反转链表中的节点，并返回修改后的链表。该算法保持节点值不变，仅改变节点指向，并且只使用常数内存。举例说明了当K为2和3时链表的变化。

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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

/**
 * 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)  
    {  
        if (!head||!head->next||k<=1)  
            return head;  
      
        ListNode* pFirst=head;  
        ListNode* pBefore=NULL;  
        ListNode* 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(ListNode* pBefore,ListNode*& pFirst,ListNode*& pKth)  
    {  
        ListNode* pN=pFirst;  
        while(pN!=pKth)  
        {  
            ListNode* pTmp=pN->next;  
            pN->next=pKth->next;  
            pKth->next=pN;  
            pN=pTmp;  
        }  
        ListNode* pTmp=pFirst;  
        pFirst=pKth;  
        pKth=pTmp;  
            if (pBefore)  
                pBefore->next=pFirst;  
    }  
};