Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

• Only constant extra memory is allowed.
• You may not alter the values in the list‘s nodes, only nodes itself may be changed.

題意大概是給定鏈表和一個參數K，每K個節點反轉一次，剩下的節點不夠K個則不反轉。

這個題目對時間復雜度要求比較高，遍歷到列表中通過reverse反轉肯定是超時的，只能在鏈表上直接操作反轉。

代碼如下：

class Solution:
    def reverseKGroup(self, head, k):
        if head==None:
            return head
        out=[]
        while True:              #遍歷鏈表，將鏈表放到一個list中，方便後續反轉指向的操作
            out.append(head)
            if head.next==None:
                break
            head=head.next
        
 
if k>len(out):
            return out[0]
        st=0
        end=k
        while True:
            for i in range(st+1,end):    #將每K個範圍內的節點指向反轉
                out[i].next=out[i-1]  
                
            if end+k<=len(out):          #判斷K範圍內最後一節點的指向，如果還存在下一個K則指向下個K的最後一個值
                out[st].next=out[end+k-1]
            elif st+k>=len(out):          #如果沒有下個K則指向None
                out[st].next=None
            elif st+k<len(out)<end+k:     #如果剩下的數不夠湊齊一個K，則指向剩下列表的第一個數
                out[st].next=out[end]
            st+=k
            end+=k  
            if len(out) < end:
                break
                
        if len(out)<2:
            return out[0]
        return out[k-1]

LeetCode算法題python解法：25. Reverse Nodes in k-Group