Leetcode题库——25.k个一组翻转链表

---

@author: ZZQ
@software: PyCharm
@file: ReverseList.py
@time: 2018/11/6 15:13
题目要求：给出一个链表，每 k 个节点一组进行翻转，并返回翻转后的链表。
k 是一个正整数，它的值小于或等于链表的长度。如果节点总数不是 k 的整数倍，那么将最后剩余节点保持原有顺序。
示例 :
给定这个链表：1->2->3->4->5
当 k = 2 时，应当返回: 2->1->4->3->5
当 k = 3 时，应当返回: 3->2->1->4->5
说明 :
你的算法只能使用常数的额外空间。
你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。
思路：针对每个子链表进行反转，需要在每次做翻转时记录当前子链表的首节点，尾节点以及下一个待翻转链表的首节点。
需要注意当剩余节点数小于k的情况以及链表本身为空的情况。

class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution():
    def __init__(self):
        pass

    def reverseKGroup(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """

        if k == 0 or k == 1 or head is None or head.next is None:
            return head

        NextListHeadNode = head
        NewH = ListNode(0)
        tempNewH = ListNode(0)
        time = 0
        while head.next is not None and tempNewH is not None:
            if time == 0:
                count, PreListFirstNode, PreListLastNode, NextListHeadNode = self.ReverseKList(head, k)
                time += 1
                NewH = PreListFirstNode
                PreListLastNode.next = NextListHeadNode
                tempNewH = NewH
                if count < k:
                    for i in range(count):
                        tempNewH = tempNewH.next
                else:
                    for i in range(count - 1):
                        tempNewH = tempNewH.next
            else:
                head = NextListHeadNode
                count, PreListFirstNode, PreListLastNode, NextListHeadNode = self.ReverseKList(head, k)
                PreListLastNode.next = NextListHeadNode
                tempNewH.next = PreListFirstNode
                time += 1
                tempNewH = tempNewH
                if count < k:
                    for i in range(count+1):
                        tempNewH = tempNewH.next
                else:
                    for i in range(count):
                        tempNewH = tempNewH.next

        return NewH

    def ReverseKList(self, head, k):

        PreH = ListNode(0)
        PostH = head.next
        count = 0
        while head is not None and count < k:
                head.next = PreH
                PreH = head
                head = PostH
                if PostH is not None:
                    PostH = PostH.next
                count += 1
        if head is None and count < k:  # 如果剩余的节点个数小于k，则返回原来的节点顺序
            lastFirstNode = self.ReverselastList(PreH)
            lastEndnode = lastFirstNode
            for i in range(count-1):
                lastEndnode = lastEndnode.next
            lastEndnode.next = None
            return count, lastFirstNode, lastEndnode, None
        else:
            tempPreH = PreH
            tt = count
            while tt-1 > 0:
                tempPreH = tempPreH.next
                tt -= 1
            return count, PreH, tempPreH, head
        # 返回当前子链表经过翻转后的首节点，尾节点以及下一个子链表未反转前的首节点

    def ReverselastList(self, head):
        PreH = ListNode(0)
        PostH = head.next
        while head is not None:
                head.next = PreH
                PreH = head
                head = PostH
                if PostH is not None:
                    PostH = PostH.next
        return PreH.next