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.

最开始想直接用双指针做就行了。 但是发现n竟然可以比length长。

/**
 * 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) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        ListNode fast = head;
        ListNode cur = head;
        if(n == 0) return head;
        if(head == null)return null;
        for(int i = 0; i < n; i ++){
            if(fast == null) return head;
            fast = fast.next;
        }
        if(fast == null) return head;
        while(fast.next != null){
            fast = fast.next;
            cur = cur.next;
        }
        ListNode hd = cur.next;
        cur.next = null;
        fast.next = head;
        return hd;
    }
}

Solution：

首先从head开始跑，直到最后一个节点，这时可以得出链表长度len。然后将尾指针指向头指针，将整个圈连起来，接着往前跑len – k%len，从这里断开，就是要求的结果了。

/**
 * 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) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
       ListNode cur = head;
       int len = 1;
       if(head == null) return null;
       while(cur.next != null){
           cur = cur.next;
           len ++;
       }
       cur.next = head;
       int t= len - n % len;
       for(int i = 0; i < t; i ++){
           cur = cur.next;
       }
       ListNode hd = cur.next;
       cur.next = null;
       return hd;
    }
}