170. Rotate List【medium】

Given a list, rotate the list to the right by k places, where k is non-negative.

 

Example

Given 1->2->3->4->5 and k = 2, return 4->5->1->2->3.

解法一：

 public class Solution {
     private int getLength(ListNode head) {
         int length = 0;
         while (head != null) {
             length ++;
             head = head.next;
         }
         return length;
     }

     public ListNode rotateRight(ListNode head, int n) {
         if (head == null) {
             return null;
         }

         int length = getLength(head);
         n = n % length;

         ListNode dummy = new ListNode(0);
         dummy.next = head;
         head = dummy;

         ListNode tail = dummy;
         for (int i = 0; i < n; i++) {
             head = head.next;
         }

         while (head.next != null) {
             tail = tail.next;
             head = head.next;
         }

         head.next = dummy.next;
         dummy.next = tail.next;
         tail.next = null;
         return dummy.next;
     }
 }

快慢指针  + dummy节点，参考@NineChapter 的代码

解法二：

 class Solution {
 public:
     /**
      * @param head: the list
      * @param k: rotate to the right k places
      * @return: the list after rotation
      */
     ListNode *rotateRight(ListNode *head, int k) {
         if (head == NULL) {
             return head;
         }

         int len = ;
         for (ListNode *node = head; node != NULL; node = node->next) {
             len++;
         }
         k = k % len;

         if (k == ) {
             return head;
         }

         ListNode *fast = head;
         for (int i = ; i < k; i++) {
             fast = fast->next;
         }

         ListNode *slow = head;
         while (fast->next != NULL) {
             slow = slow->next;
             fast = fast->next;
         }

         fast->next = head;
         head = slow->next;
         slow->next = NULL;

         return head;
     }
 };

不带dummy节点的解法，参考@NineChapter 的代码