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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：

 1 public class Solution {
 2     private int getLength(ListNode head) {
 3         int length = 0;
 4         while (head != null) {
 5             length ++;
 6             head = head.next;
 7         }
 8         return length;
 9     }
10     
11     public ListNode rotateRight(ListNode head, int n) {
12         if (head == null) {
13             return null;
14         }
15         
16         int length = getLength(head);
17         n = n % length;
18         
19         ListNode dummy = new ListNode(0);
20         dummy.next = head;
21         head = dummy;
22         
23         ListNode tail = dummy;
24         for (int i = 0; i < n; i++) {
25             head = head.next;
26         }
27         
28         while (head.next != null) {
29             tail = tail.next;
30             head = head.next;
31         }
32         
33         head.next = dummy.next;
34         dummy.next = tail.next;
35         tail.next = null;
36         return dummy.next;
37     }
38 }

快慢指针  + dummy节点，参考@NineChapter 的代码

 

解法二：

 1 class Solution {
 2 public:
 3     /**
 4      * @param head: the list
 5      * @param k: rotate to the right k places
 6      * @return: the list after rotation
 7      */
 8     ListNode *rotateRight(ListNode *head, int k) {
 9         if (head == NULL) {
10             return head;
11         }
12         
13         int len = 0;
14         for (ListNode *node = head; node != NULL; node = node->next) {
15             len++;
16         }
17         k = k % len;
18         
19         if (k == 0) {
20             return head;
21         }
22         
23         ListNode *fast = head;
24         for (int i = 0; i < k; i++) {
25             fast = fast->next;
26         }
27         
28         ListNode *slow = head;
29         while (fast->next != NULL) {
30             slow = slow->next;
31             fast = fast->next;
32         }
33         
34         fast->next = head;
35         head = slow->next;
36         slow->next = NULL;
37         
38         return head;
39     }
40 };

不带dummy节点的解法，参考@NineChapter 的代码

 

posted on 2017-12-31 11:40  LastBattle  阅读(101)  评论(0)  编辑  收藏  举报