Rotate List 旋转链表 @LeetCode

链表的循环，切割问题

package Level3;

import Utility.ListNode;

/**
 * 
 * Rotate List
 * 
 * Given a list, rotate the list to the right by k places, where k is
 * non-negative.
 * 
 * For example: Given 1->2->3->4->5->NULL and k = 2, return 4->5->1->2->3->NULL.
 * 
 */
public class S61 {

	public static void main(String[] args) {
		ListNode head = new ListNode(1);
		ListNode n2 = new ListNode(2);
		ListNode n3 = new ListNode(3);
		ListNode n4 = new ListNode(4);
		ListNode n5 = new ListNode(5);
		head.next = n2;
		n2.next = n3;
		n3.next = n4;
		n4.next = n5;
		
		ListNode rotateHead = rotateRight(head, 2);
		rotateHead.print();
	}

	public static ListNode rotateRight(ListNode head, int n) {
		if(head == null){
			return null;
		}
		ListNode back = head;
		ListNode front = head;
		ListNode end = head;
		
		// 先把链表改为循环链表
		while(front.next != null){
			front = front.next;
		}
		end = front;		// 记录原尾节点
		front.next = head;	// 形成环
		front = head;		// 复原front指针
		
		// 使得front在back前面n个距离
		for(int i=0; i<n; i++){
			front = front.next;
		}
		
		// 双指针同步移动
		while(front != end){
			front = front.next;
			back = back.next;
		}
		
		ListNode rotateHead = back.next;		// 找到rotate之后的链表头
		back.next = null;			// 切开循环链表
		return rotateHead;
	}

}

要点就是先把单链表变成循环链表

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode rotateRight(ListNode head, int n) {
        if(head == null) {
            return null;
        }
        ListNode cur = head;
        while(cur.next != null) {
            cur = cur.next;
        }
        cur.next = head;
        
        ListNode fast = head, slow = head;
        for(int i=0; i<n; i++) {
            fast = fast.next;
        }
        while(fast.next != head){
            fast = fast.next;
            slow = slow.next;
        }
        ListNode newHead = slow.next;
        slow.next = null;
        return newHead;
    }
}