Question:

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.

 

Analysis:

给出一个单链表，旋转链表的从右边数k个位置，将他们按顺序放到左边。k是个非负的整数。

注意一些特殊情况：

如当给出的k值比链表长度还大怎么办？当k=0怎么办？

 

Answer：

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode rotateRight(ListNode head, int k) {        if(head == null || head.next == null)             return head;        int len = 0;        ListNode p = head;        ListNode end = null;        while(p != null) {            if(p.next == null)                 end = p;            p = p.next;            len++;        }        k %= len;        if(k == 0)            return head;        int pre = len - k;        len = 0;        p = head;        while(len +1 < pre) {            p = p.next;            len++;        }        ListNode q = p.next;        end.next = head;        p.next = null;        return q;    }}