【leetcode刷题】20T27-旋转链表

示例 1:
输入: 1->2->3->4->5->NULL, k = 2
输出: 4->5->1->2->3->NULL
解释:
向右旋转 1 步: 5->1->2->3->4->NULL
向右旋转 2 步: 4->5->1->2->3->NULL

示例 2:
输入: 0->1->2->NULL, k = 4
输出: 2->0->1->NULL
解释:
向右旋转 1 步: 2->0->1->NULL
向右旋转 2 步: 1->2->0->NULL
向右旋转 3 步: 0->1->2->NULL
向右旋转 4 步: 2->0->1->NULL

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def rotateRight(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if k == 0 or head is None:
            return head
        # 统计节点个数
        p = head
        count = 1
        while p.next is not None:
            p = p.next
            count += 1

        # k对count取余
        k %= count
        if k == 0:
            return head

        # 找到旋转位置
        count = count - k - 1
        q = head
        while count > 0:
            count -= 1
            q = q.next

        # 旋转
        p.next = head
        head = q.next
        q.next = None
        return head