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 用双指针找到倒数第n个ListNode的位置,也就是旋转的节点。

这个题只需要注意一下记录链表长度。

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def rotateRight(self, head, k):        """        :type head: ListNode        :type k: int        :rtype: ListNode        """        if head is None:return None        fast = slow = head               length = 0        while fast.next:            fast = fast.next            length += 1                k = k%(length+1)        fast = head        while k>0 and fast.next:            fast = fast.next            k -= 1        while fast.next and slow.next:            fast = fast.next            slow = slow.next        if slow.next is None:            return head        nhead = p = slow.next        slow.next = None        while p.next:            p=p.next        p.next=head        return nhead

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