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LeetCode.旋转链表

发表于 分类于 Valine：

给定一个链表，旋转链表，将链表每个节点向右移动 k 个位置，其中 k 是非负数。

  • 示例 1:
    输入: 1->2->3->4->5->NULL, k = 2
    输出: 4->5->1->2->3->NULL
  • 示例 2:
    输入: 0->1->2->NULL, k = 4
    输出: 2->0->1->NULL

解法：

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object LinkedListMove {
     def main(args: Array[String]): Unit = {
          println(
               rotateRight(ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5, null))))), 0)
          )
          println(
               rotateRight(ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5, null))))), 1)
          )
          println(
               rotateRight(ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5, null))))), 2)
          )
          println(
               rotateRight(ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5, null))))), 3)
          )
          println(
               rotateRight(ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5, null))))), 4)
          )
          println(
               rotateRight(ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5, null))))), 5)
          )
          println(
               rotateRight(ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5, null))))), 6)
          )
     }

     def length(node: ListNode): Int = {
          var head = node
          var len = 0
          while (head != null) {
               len += 1
               head = head.next
          }
          len
     }

     def rotateRight(node: ListNode, k: Int): ListNode = {
          if (k == 0) return node
          if (node == null) return null
          if (node.next == null) return node
          val len = length(node)
          var newk = k % len
          if (newk == 0) return node

          newk = len - newk
          val start = node
          var newHead = node

          for (i <- 0 until newk - 1) {
               newHead = newHead.next
          }

          var t = newHead.next
          newHead.next = null
          newHead = t
          while (t.next != null) {
               t = t.next
          }

          t.next = start
          newHead
     }

     case class ListNode(x: Int, var next: ListNode)

}