【LeetCode】 Rotate List 循环链表

题目：rotate list

解法1：

<span style="font-size:18px;">/**LeetCode Rotate List：Given a list, rotate the list to the right by k places, where k is non-negative.
 * 题目：循环移动链表，等价于将链表从右边数的k个节点移动到表的前方
 * 思路：移动倒是简单。重点是要找到链表倒数的k个数，就等价于找到倒数第K+1个数，设置两个指针，先后遍历链表。中间相隔k个数
 * 当前面的指针走到最后的一个节点，此时后面的指针指向的就是倒数第k+1个数
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */ 
package javaTrain;

public class Train14 {
	 public ListNode rotateRight(ListNode head, int k) {
	        ListNode pFast,pSlow,pKnode;
	        int n = 0;
	        
	        if(head == null || k < 1 ) return head;	//注意特殊情况
	        pFast = head;
	        pSlow = head;
	        while(pFast != null){
	        	pFast = pFast.next;
	        	n++;
	        }
	        k = k%n;	//循环移动，能够转变为求模
	        if(k == 0) return head;	//移动的次数等于自己的长度。等价于本身
	        pFast = head;
	        while(k>0 && pFast != null){ 
	        	pFast = pFast.next;
	        	k--;
	        } 
	        while(pFast.next != null){
	        	pFast = pFast.next;
	        	pSlow = pSlow.next;
	        }
	        pKnode = pSlow.next;	//第k+1个节点，次后就是要移到前面的节点了，
	        pSlow.next = null;
	        pFast.next = head;	//原本最后的节点此时排在头结点之前
	        
	        return pKnode;
	    } 
}
</span>

解法2：

<span style="font-size:18px;">//法2：将链表连城环，而后从新寻找新的头结点和尾节点，即在len-k处
package javaTrain;

public class Train14_1 {
	public ListNode rotateRight(ListNode head, int k) {
		if(head == null || k == 0) return head;	//特殊情况
		ListNode pNode = head;
		int len = 1;
		while(pNode.next != null){
			pNode = pNode.next;
			len++;
		}
		k = len-k%len;
		pNode.next = head;	//注意此时pNode是原来的尾节点
		for(int i = 0;i < k;i++){
			pNode = pNode.next;		
		}
		head = pNode.next;
		pNode.next = null;
		return head;
	}
}
</span>