Given a list, rotate the list to the right by k places, where k is non-negative.
For example: Given1->2->3->4->5->NULLand k =2,
return4->5->1->2->3->NULL.
我的想法是找到要倒置的第一个元素位置，然后再由末尾元素指向head。
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode rotateRight(ListNode head, int n) { if(head == null || head.next == null) return head; ListNode cur = head; int count = 1; while(cur.next != null) { cur = cur.next; count++; } ListNode p = head; ListNode pre = new ListNode(0); pre.next = p; int s = count - n; while(s != 0){ p = p.next; pre = pre.next; s--; } cur.next = head; pre.next = null; return p; } }运行是如下结果：
查看别人通过的代码：
public class Solution { public ListNode rotateRight(ListNode head, int k) { if(head==null||k==0) return head; int len = 0; ListNode cur = head; ListNode last = null; while(cur!=null){ len++; last = cur; cur = cur.next; } if(k>=len&&len!=0) k %= len; if(k==0) return head; last.next = head;//形成环路 cur = head;//因为是右移动，所以从头开始遍历到倒数第k个元素 for(int i = 1;i<=len-k;i++){ last = cur; cur = cur.next; } //cur是头 last.next = null; return cur; } } 好像没啥区别。。。