[LeetCode][Java] Reverse Nodes in k-Group

题目：

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

题意：

给定一个链表，对于链表中的节点每k个进行转置一次，返回改动后的链表。

如果链表中的节点的数目不是k的整数倍，那么保持结尾剩余的那些节点不懂。

只允许常数空间。

比如：

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

算法分析：

利用的题目《Reverse Linked List II》的算法，k个元素为一组，进行倒置 

AC代码：

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */ //利用的题目《Reverse Linked List II》的算法，k个元素为一组，进行倒置 public class Solution {    public ListNode reverseKGroup(ListNode head, int k)     {if(head==null||k<=1) return head;    int startindex=1;int endindex=k;int ki=0;int Allk=0;int kk=0;ListNode testhead=head;while(testhead!=null){kk++;testhead=testhead.next;}Allk=kk/k;while(ki<Allk){head=reverseBetween( head, startindex, endindex); ki++;startindex+=k;endindex+=k;}    return head;    }    public static ListNode reverseBetween(ListNode head, int m, int n)     {        if(head == null)            return null;        ListNode dummy = new ListNode(0);        dummy.next = head;        ListNode preNode = dummy;        int i=1;        while(preNode.next!=null && i<m)        {            preNode = preNode.next;            i++;        }        //if(i<m)          //  return head;        ListNode mNode = preNode.next;        ListNode cur = mNode.next;        while(cur!=null && i<n)        {            ListNode next = cur.next;            cur.next = preNode.next;            preNode.next = cur;            mNode.next = next;            cur = next;            i++;        }        return dummy.next;    }}