25.leetcode 25 Reverse Nodes in k-Group

本題是一個hard鏈表翻轉問題，關鍵點如下：

1.子鏈表翻轉，需要有一個頭節點指引返回結果的head

2.題目有空間要求，合適的做法是先計數再做子鏈表翻轉

代碼如下：

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def reverseSubList(self, currN, groupL):
        preN = None
        while groupL:
            nextN = currN.next
            currN.next = preN
            preN = currN
            currN = nextN
            groupL -= 1
        return preN
            
    def reverseKGroup(self, head, k):
        end = head
        dummy = ListNode(0) #head之前的節點
        result = dummy
        length = 1
        while end:
            end = end.next
            if length != k:
                length += 1
            else:
                dummy.next = self.reverseSubList(head, k) #返回已經倒序的子鏈的head,被前一個節點指引
                dummy = head #head是本子鏈第一個節點，倒敘後爲下個子鏈的head前一個節點，傳給dumy
                head = end   #end爲下個子鏈第一個節點，給head
                length = 1
        dummy.next = head #若後面的子鏈長度小於k，則直接傳給dummy，否則此處傳的是None
        return result.next