问题：
方法：
第一遍遍历先获得链表长度，然后通过取余去除无效的旋转次数，得到有效次数j。然后移动链表到倒数第j个点，把该点作为新的链表的头节点，然后把原来的头节点接到新链表的尾部即可。
package com.eric.leetcode
class RotateList {
fun rotateRight(head: ListNode?, k: Int): ListNode? {
var cur = head
var count = 1
while (cur?.next != null) {
count++
cur = cur.next
}
val end = cur
cur = head
val j = k % count
for (index in 1 until (count - j)) {
cur = cur?.next
}
end?.next = head
val result = cur?.next
cur?.next = null
return result
}
}
有问题随时沟通
