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[LeetCode] 61. Rotate List

Given the head of a linked list, rotate the list to the right by k places.

Example 1:

Input: head = [1,2,3,4,5], k = 2
Output: [4,5,1,2,3]

Example 2:

Input: head = [0,1,2], k = 4
Output: [2,0,1]

Constraints:

The number of nodes in the list is in the range [0, 500].
-100 <= Node.val <= 100
0 <= k <= 2 * 109

旋转链表。

题意很简单，给一个linked list和一个数字k，请你对linked list向右rotate K次，输出rotate后的结果。因为是单链表所以没法从右往左遍历链表。思路是将input先连成一个环，然后在新的head节点和其之前一个节点处断开即可。代码里面注意几个细节。

1. 算K的长度的时候，注意有可能K是大于链表长度len的，需要取余；

2. 22行的for loop是遍历到新的head节点之前的那个节点，然后再将其跟新的head断开；

跑一个例子

1 - 2 - 3 - 4 - 5, k = 2

先移动到链表末端，得到长度len = 5

1 - 2 - 3 - 4 - 5

|______________|

h

h

将链表末端再连到head节点，形成一个环

因为需要往右rotate 2次，所以实际上是要将head指针往后移动len - k次，这样head指针会在新的head之前的一个节点上（4）。此时新的head节点是head.next

最后将4和5之间的连接断开即可

时间O(n)

空间O(1)

JavaScript实现

 1 /**
 2  * @param {ListNode} head
 3  * @param {number} k
 4  * @return {ListNode}
 5  */
 6 var rotateRight = function(head, k) {
 7     // corner case
 8     if (head == null || head.next == null) {
 9         return head;
10     }
11 
12     // normal case
13     let len = 1;
14     let index = head;
15     while (index.next !== null) {
16         index = index.next;
17         len++;
18     }
19     index.next = head;
20 
21     // loop again
22     for (let i = 1; i < len - (k % len); i++) {
23         head = head.next;
24     }
25     let res = head.next;
26     head.next = null;
27     return res;
28 };

Java实现

 1 class Solution {
 2     public ListNode rotateRight(ListNode head, int k) {
 3         // corner case
 4         if (head == null || head.next == null) {
 5             return head;
 6         }
 7         // get the length
 8         ListNode index = head;
 9         int len = 1;
10         while (index.next != null) {
11             index = index.next;
12             len++;
13         }
14 
15         index.next = head;
16         for (int i = 1; i < len - k % len; i++) {
17             head = head.next;
18         }
19         ListNode res = head.next;
20         head.next = null;
21         return res;
22     }
23 }