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.
一刷
题解：
两种思路，先遍历一遍求出链表的长度len, 用k-len得到距离头部的距离，然后找到需要reverse的节点的前一个节点。
/**
* 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 || k<0) return null;
ListNode node = head;
int len = 1;
while(node.next!=null){
node = node.next;
len++;
}
node.next = head;
k = len-k %len;
while(k>0){
node = node.next;
k--;
}
head = node.next;
node.next = null;
return head;
}
}
第二个思路，快慢指针， 得到需要reverse为head的前一个节点.
二刷：
与一刷思路相同，先求得整个链表的长度len，避免k过大。真正的k为k%len. 然后求得第i-k个点（表头为1），那么该点的next需要被移至head
public ListNode rotateRight(ListNode head, int n) {
if (head==null||head.next==null) return head;
ListNode dummy=new ListNode(0);
dummy.next=head;
ListNode fast=dummy,slow=dummy;
int i;
for (i=0;fast.next!=null;i++)//Get the total length
fast=fast.next;
for (int j=i-n%i;j>0;j--) //Get the i-n%i th node
slow=slow.next;
fast.next=dummy.next; //Do the rotation
dummy.next=slow.next;
slow.next=null;
return dummy.next;
}
