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.
我是利用双指针吧,就可以在一次遍历的情况下找到要断开的位置即倒数第几个位置,然后插到头上即可。
/**
* 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 (k == 0 || head == null || head.next == null)
return head;
ListNode p, q;
p = q = head;
int len = 1;
while (p.next != null) {
p = p.next;
len++;
}
k = k % len;
if (k == 0)
return head;
p = head;
for (int i = 0; i < k; i++) {
p = p.next;
}
while (p.next != null) {
q = q.next;
p = p.next;
}
p.next = head;
p = q.next;
q.next = null;
head = p;
return head;
}
}