Reverse Nodes in k-Group 指针操作 每k个翻转链表

Reverse Nodes in k-Group

Given a linked list, reverse the nodes of a linked list k

If the number of nodes is not a multiple of k

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:
//将链表按每K个进行逆转，不够k个时，不变。
    ListNode* reverseKGroup(ListNode* head, int k) {
        
        int len=0;
        ListNode *p=head;
        while(p)
        {
            len++;
            p=p->next;
        }
        if(len==0 || k==1 || k>len)
            return head;
            
        ListNode *help=new ListNode(0);
        help->next=head;
        
        ListNode *reverseHead=help;
        
        p=head;
        while(k<=len)
        {
            int tmp=k;
            ListNode *mark=p;
            
            ListNode *pre;
            while(tmp>0)
            {
                pre=p;
                p=p->next;
                pre->next=help->next;
                help->next=pre;
                tmp--;
            }
            mark->next=p;
            help=mark;
            
            len=len-k;
        }
        
        return reverseHead->next;
    }
};