61. Rotate List
摘要：
Given a linked list, rotate the list to the right by k places, where k is non-negative.
Example 1:
Input: 1->2->3->4->5->NULL, k...
Given a linked list, rotate the list to the right by k places, where k is non-negative.
Example 1:
Input: 1->2->3->4->5->NULL, k = 2 Output: 4->5->1->2->3->NULL Explanation: rotate 1 steps to the right: 5->1->2->3->4->NULL rotate 2 steps to the right: 4->5->1->2->3->NULL
Example 2:
Input: 0->1->2->NULL, k = 4 Output: 2->0->1->NULL Explanation: rotate 1 steps to the right: 2->0->1->NULL rotate 2 steps to the right: 1->2->0->NULL rotate 3 steps to the right: 0->1->2->NULL rotate 4 steps to the right: 2->0->1->NULL
難度：medium
題目：給定連結串列，向右旋轉k個位置，k為非負整數。
思路：首先統計結點個數，並記錄尾結點。然後用前後指標找出倒數第k + 1個結點。
Runtime: 7 ms, faster than 92.47% of Java online submissions for Rotate List.
Memory Usage: 27 MB, less than 41.82% of Java online submissions for Rotate List.
/**
* Definition for singly-linked list.
* public class ListNode {
*int val;
*ListNode next;
*ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode rotateRight(ListNode head, int k) {
if (null == head || k <= 0) {
return head;
}
int cnt = 0;
ListNode ptr = head,tail = ptr;
while (ptr != null) {
tail = ptr;
ptr = ptr.next;
cnt++;
}
k = k % cnt;
ListNode dummyHead = new ListNode(0), lastKPtr = dummyHead;
dummyHead.next = head;
ptr = dummyHead.next;
while (ptr != null) {
if (--k < 0) {
lastKPtr = lastKPtr.next;
}
ptr = ptr.next;
}
tail.next = dummyHead.next;
dummyHead.next = lastKPtr.next;
lastKPtr.next = null;
return dummyHead.next;
}
}