【LeetCode】25.Reverse Nodes in k-Group

所有题目索引

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

将链表的每k个分为一组，组内反转

思路

类似于上一题，递归解答，每次反转一组的内容

代码

java实现

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
int count=0;
ListNode nextH=head;
while(nextH!=null&&count!=k)
{
nextH=nextH.next;
count++;
}
if(count==k)
{
ListNode cur=head;
ListNode nxt=head.next;
head.next=reverseKGroup(nextH,k);
while(nxt!=nextH)
{
ListNode pre=cur;
cur=nxt;
nxt=nxt.next;
cur.next=pre;
}
head=cur;
}
return head;
}
}