题目：
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.
链接：https://leetcode.com/problems/rotate-list/#/description
4/26/2017
19ms， 35%
先找到链表的长度，并找到相对移动的k
用快指针找到距离差，用快慢指针一起找到新的head
注意
1. 判断nodeVisited之后还有判断k在模length之后是否为0，0的话可以直接返回
2. 第19行，fast应该重新回到head
1 public class Solution { 2 public ListNode rotateRight(ListNode head, int k) { 3 if (head == null) return null; 4 5 int length, nodeVisited = k; 6 ListNode slow = head, fast = head; 7 8 while (fast != null && nodeVisited > 0) { 9 fast = fast.next; 10 nodeVisited--; 11 } 12 if (nodeVisited == 0 && fast == null) return head; 13 else if (nodeVisited > 0) { 14 length = k - nodeVisited; 15 k = k % length; 16 } 17 if (k == 0) return head; 18 19 fast = head; 20 while (k > 0) { 21 fast = fast.next; 22 k--; 23 } 24 25 while (fast.next != null) { 26 slow = slow.next; 27 fast = fast.next; 28 } 29 ListNode temp = slow.next; 30 slow.next = null; 31 fast.next = head; 32 return temp; 33 34 } 35 }
别人的算法：
变成环，然后再来找
1 class Solution { 2 public: 3 ListNode* rotateRight(ListNode* head, int k) { 4 if(!head) return head; 5 6 int len=1; // number of nodes 7 ListNode *newH, *tail; 8 newH=tail=head; 9 10 while(tail->next) // get the number of nodes in the list 11 { 12 tail = tail->next; 13 len++; 14 } 15 tail->next = head; // circle the link 16 17 if(k %= len) 18 { 19 for(auto i=0; i<len-k; i++) tail = tail->next; // the tail node is the (len-k)-th node (1st node is head) 20 } 21 newH = tail->next; 22 tail->next = NULL; 23 return newH; 24 } 25 };
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