题目

给你一个链表的头节点 head ，旋转链表，将链表每个节点向右移动 k 个位置。
示例 1：
输入：head = [1,2,3,4,5], k = 2
输出：[4,5,1,2,3]

分析

这道题目用快慢指针即可解，先快指针往前走k个位置，然后和慢指针同时开始遍历，直到快指针的next为空。这个时候慢指针的next就是最新的头结点。这里要注意k比链表总长度大的情况，可以通过求余解决

public class LinkNode {int val;LinkNode next;public LinkNode(int data) {this.val = data;this.next = null;}
}
public class LinkList {LinkNode head;public LinkList() {this.head = null;}public LinkNode getHead() {return this.head;}//添加元素public void addNode(int data) {LinkNode node = new LinkNode(data);if (this.head == null) {this.head = node;} else {LinkNode cur = this.head;while(cur.next != null) {cur = cur.next;}cur.next = node;}}//正序打印public void print(LinkNode node) {while(node != null) {System.out.print(node.val);System.out.print(" ");node = node.next;}System.out.println();}public void rotate(int k) {if(this.head == null) {return;}LinkNode p = this.head;int cnt = 0;while(p!= null) {cnt++;p = p.next;}k = k % cnt;LinkNode fast = this.head;LinkNode slow = this.head;for(int i = 0;i<k;i++) {fast = fast.next;}while(fast.next != null) {fast = fast.next;slow = slow.next;}fast.next = head;fast = slow.next;slow.next = null;print(fast);}
}public class rotateList {public static void main(String[] args) {LinkList list = new LinkList();list.addNode(1);list.addNode(2);list.addNode(3);list.addNode(4);list.addNode(5);list.rotate(2);}
}