[Leetcode] Reverse Nodes in k-Group (Java)

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

链表逆序，遍历时把遍历到的node放在首位，最后一个就成为了第一个，第一个成为最后一个

public class ReverseNodesinkGroup {static class ListNode {int val;ListNode next;ListNode(int x) {val = x;next = null;}}public ListNode reverse(ListNode pre,ListNode next){ListNode last = pre.next;ListNode cur = last.next;while(cur!= next){last.next = cur.next;cur.next = pre.next;pre.next = cur;cur = last.next;}return last;}public ListNode reverseKGroup(ListNode head, int k) {if(k<=1||head==null)return head;ListNode res = new ListNode(-1);res.next = head;int i=0;ListNode pre = res;while(head!=null) {i++;if(i%k==0){pre = reverse(pre, head.next);head = pre.next;}else {head = head.next;}}return res.next;}public static void main(String[] args) {ListNode l1 = new ListNode(1);ListNode l2 = new ListNode(2);ListNode l3 = new ListNode(3);ListNode l4 = new ListNode(4);ListNode l5 = new ListNode(5);l1.next=l2;l2.next=l3;l3.next=l4;l4.next=l5;int k = 3;ListNode res = new ReverseNodesinkGroup().reverseKGroup(l1, k);if(res!=null){System.out.print(res.val);res=res.next;}while(res!=null){ System.out.print("->"+res.val);res=res.next;}}}