25. Reverse Nodes in k-Group leetcode Python 2016 new season-优快云博客

本文链接：https://blog.youkuaiyun.com/hyperbolechi/article/details/50502741

本文介绍了一种算法，该算法将链表中的节点每K个一组进行反转，并返回修改后的链表。讨论了特殊情况处理及实现细节，提供了一个O(n)时间复杂度的解决方案。

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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

need a helper function to reverse k nodes and then put the reversed list together. this should be O(n) time.

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

class Solution(object):
    def reverseNodes(self, startNode, endNode):
        dummyNode = ListNode(None)
        dummyNode.next = startNode
        while dummyNode.next != endNode:
            tmpNode = startNode.next
            startNode.next = tmpNode.next
            tmpNode.next = dummyNode.next
            dummyNode.next = tmpNode
        return endNode, startNode
    def reverseKGroup(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if not head:
            return head
        dummyNode = ListNode(None)
        dummyNode.next = head
        startNode = head
        while startNode.next:
            endNode = startNode
            for count in range(k - 1):
                endNode = endNode.next
                if not endNode.next:
                    return dummyNode
            newstart, newend = self.reverseNodes(startNode, endNode)
            startNode.next = newstart
            startNode = newend
        return dummyNode.next