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.
此题先想需要多少个ListNode，dummy，用于返回。fast，slow。三个就够了，我开始做的时候，多用了一个cur来计算链表长度，我的原始代码如下：
/**
* 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) {
ListNode slow = head;
ListNode fast = head;
ListNode cur = head;
ListNode dummy = new ListNode(0);
if(head==null) return head;
int len = 0;
while(cur!=null){
cur=cur.next;
len++;
}
k = k%len;
if(k==0) return head;
while(k>0){
fast = fast.next;
k--;
}
while(fast!=null&&fast.next!=null){
fast = fast.next;
slow = slow.next;
}
fast.next = head;
dummy.next = slow.next;
slow.next = null;
return dummy.next;
}
}
看了答案以后，发现答案的方法很简明，答案代码如下：
/**
* 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) {
ListNode dummy = new ListNode(0);
ListNode slow = dummy;
ListNode fast = dummy;
dummy.next = head;
if(head==null||head.next==null) return head;
int len = 0;
while(fast.next!=null){
fast = fast.next;
len++;
}
k = k%len;
while(len-k>0){
slow = slow.next;
len--;
}
fast.next = dummy.next;
dummy.next = slow.next;
slow.next = null;
return dummy.next;
}
}