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 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://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;    }};