标签：style blog io color sp for div on log
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.
Solution 1:
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。
这个既然要rotate，不如先连起来loop，这样怎么也不怕null pointer exception了。然后再找到该断开的地方断开。
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 public ListNode rotateRight(ListNode head, int n) { 14 if(head==null||head.next==null) 15 return head; 16 ListNode temp=head; 17 int count=1; 18 while(temp.next!=null){ 19 count++; 20 temp=temp.next; 21 } 22 temp.next=head; 23 n%=count; 24 ListNode p=head; 25 for(int i=0;i<count-n-1;i++){ 26 p=p.next; 27 } 28 temp=p.next; 29 p.next=null; 30 return temp; 31 } 32 }
Solution 2:
利用快慢指针来做。
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 public ListNode rotateRight(ListNode head, int n) { 14 if(head==null||head.next==null) 15 return head; 16 ListNode dummy=new ListNode(-1); 17 dummy.next=head; 18 ListNode fast=dummy; 19 ListNode slow=dummy; 20 ListNode temp=head; 21 int count=1; 22 while(temp.next!=null){ 23 count++; 24 temp=temp.next; 25 } 26 n%=count; 27 while(fast.next!=null&&n!=0){ 28 fast=fast.next; 29 n--; 30 } 31 if(n!=0||fast.next==null) 32 return dummy.next; 33 while(fast.next!=null){ 34 fast=fast.next; 35 slow=slow.next; 36 } 37 fast.next=dummy.next; 38 dummy.next=slow.next; 39 slow.next=null; 40 41 return dummy.next; 42 } 43 }
标签：style blog io color sp for div on log
原文地址：http://www.cnblogs.com/Phoebe815/p/4101484.html
