[leetcode][25]Reverse Nodes in k-Group

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本文介绍了一种算法，用于每K个节点反转一次链表，并返回修改后的链表。该算法不改变节点值，仅调整节点顺序，并且只使用常量内存。提供了Python实现示例。

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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.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For 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

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def reverseKGroup(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        current = head
        k_container = []
        pre_k_node = None
        while current:
            next_node = current.next
            k_container.append(current)
            if len(k_container) == k:
                reverse_k_nodes = k_container[::-1]
                if pre_k_node is None:
                    for i in range(len(reverse_k_nodes)-1):
                        reverse_k_nodes[i].next = reverse_k_nodes[i+1]
                    head = reverse_k_nodes[0]
                    pre_k_node = reverse_k_nodes[-1]
                    
                else:
                    pre_k_node.next = reverse_k_nodes[0]
                    for i in range(len(reverse_k_nodes)-1):
                        reverse_k_nodes[i].next = reverse_k_nodes[i+1]
                    pre_k_node = reverse_k_nodes[-1]
                k_container = []
            current = next_node
        
        if len(k_container) == 0:
            if pre_k_node is None:
                return None
            else:
                pre_k_node.next = None
        else:
            if pre_k_node is None:
                return k_container[0]
            else:
                pre_k_node.next = k_container[0]
        return head