leetcode -- Rotate List -- 重点

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https://leetcode.com/problems/rotate-list/

这里思路很简单。但是要注意，k可以大于len(linkedlist). 要求得长度之后取mod。还有就是这个rotate，不需要把rotate部分逆置。

    def rotateRight(self, head, k):        """        :type head: ListNode        :type k: int        :rtype: ListNode        """        if not head or not head.next: return head        #calculate the length        p = head        count = 0        while p:            p = p.next            count += 1        k = k % count        #print (count,k)        if k == 0: return head        i,j = head,head        pre_i = None        for x in xrange(k-1):              j = j.next        while j.next:            pre_i = i            i, j = i.next, j.next        if i == head:            return head        else:            new_head, new_end = i,j            new_end.next = head            pre_i.next = None            return new_head

由于一开始看错题目，以为rotate的部分要逆置。所以这里还是post出逆置的code。

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def reverse(self, head):        org_head = head        cur = head.next        pre = head        pre.next = None        while cur:            tmp = cur.next            cur.next = pre            pre, cur =cur, tmp        return pre, org_head    def rotateRight(self, head, k):        """        :type head: ListNode        :type k: int        :rtype: ListNode        """        if not head or not head.next: return head        #calculate the length        p = head        count = 0        while p:            p = p.next            count += 1        k = k % count        print (count,k)        if k == 0: return head        i,j = head,head        pre_i = None        for x in xrange(k-1):              j = j.next        while j.next:            pre_i = i            i, j = i.next, j.next        if i == head:            return self.reverse(i)[0]        else:            new_head, new_end = self.reverse(i)            new_end.next = head            pre_i.next = None            return new_head

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