Reverse Nodes in K-Groups Leetcode Python-优快云博客

本文介绍了一种算法，该算法将给定的链表每K个节点进行一次翻转，并保持剩余节点不变。文章提供了详细的实现步骤及Python代码示例。

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

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

这题做法与reverse linklist和 reverse pair类似都需要有一个用来reverse的函数。

题目的意思是每k个node翻转一次，如果不够k个就不转。

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

class Solution:
    # @param head, a ListNode
    # @param k, an integer
    # @return a ListNode
    def reverse(self,start,end):
        dummy=ListNode(0)
        dummy.next=start
        while dummy.next!=end:
            tmp=start.next
            start.next=tmp.next
            tmp.next=dummy.next
            dummy.next=tmp
        return end,start
        
    def reverseKGroup(self, head, k):
        if head==None:return None
        dummy=ListNode(0)
        dummy.next=head
        start=dummy
        while start.next:
            end=start
            for i in range(k-1):
                end=end.next
                if end.next==None:
                    return dummy.next
            newstart,newend=self.reverse(start.next,end.next)
            start.next=newstart
            start=newend
        return dummy.next