今天和大家聊的问题叫做K 个一组翻转链表，我们先来看题面：
https://leetcode-cn.com/problems/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. k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
题意
给你一个链表，每 k 个节点一组进行翻转，请你返回翻转后的链表。k 是一个正整数，它的值小于或等于链表的长度。如果节点总数不是 k 的整数倍，那么请将最后剩余的节点保持原有顺序。
样例
给你这个链表：1->2->3->4->5 当 k = 2 时，应当返回: 2->1->4->3->5 当 k = 3 时，应当返回: 3->2->1->4->5
题解
每k个节点进行一次反转，用两个指针记录k个区间的首尾节点，写一个reverse函数反转该局部内的节点指向，接着继续向后走，不停的取k个进行反转，如果不够k个就返回。具体代码如下：
import java.util.Scanner; public class Main { //定义Node节点 static class ListNode { public int val; public ListNode next = null; public ListNode(int val) { this.val = val; } } public static void main(String[] args) { //1.获取输入信息 Scanner scanner = new Scanner(System.in); String string = scanner.nextLine(); int k = scanner.nextInt(); String[] strings = string.split(" "); //2.创建头结点 ListNode head = new ListNode(0); ListNode tail = head; //3.将输入的字符串变为链表节点 for (String str : strings) { ListNode newNode = new ListNode(Integer.valueOf(str)); tail.next = newNode; tail = tail.next; } head = head.next; //每k个反转链表 ListNode node = reverseGroup(head, k); while(node!=null){ System.out.print(node.val+" "); node = node.next; } } //不停地取k个进行翻转，如果不够k个，就直接返回,结束 public static ListNode reverseGroup(ListNode head, int k) { if (head == null || head.next == null || k <= 1) return head; ListNode currentNode = head; //获取k个元素的首尾节点 for (int count = 1; count < k; count++) { currentNode = currentNode.next; //不够K个则返回 if(currentNode==null) return head; } ListNode next = currentNode.next; //对局部链表进行反转 reverse(head,currentNode); head.next=reverseGroup(next,k); return currentNode; } //写一个头尾节点反转的局部函数 public static ListNode reverse(ListNode head, ListNode tail) { if (head == null || head.next == null) return head; ListNode pre = null; ListNode next = null; while (pre != tail) { next = head.next; head.next = pre; pre = head; head = next; } return pre; } }
好了，今天的文章就到这里，如果觉得有所收获，请顺手点个在看或者转发吧，你们的支持是我最大的动力。
