【leetcode】-25. Reverse Nodes in k-Group 反转k组数组

题目

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.

递归

题目要求将小于链表长度的k个链表进行反转，不足k个的链表顺序不变。这题乍看起来很复杂，其实可以先翻转第一组k个链表，然后将剩余的部分看作一个新的链表进行递归计算。

python 代码

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        if not head:
            return None
        a = b = head
        for i in range(k):
            if not b:
                return head
            b = b.next
        newHead =  self.reverse(a,b)
        a.next = self.reverseKGroup(b,k)
        
        return newHead
    
    def reverse(self,a,b):
        pre = None
        cur = a
        nxt = a
        while cur != b:
            nxt = cur.next
            cur.next = pre
            pre = cur
            cur = nxt
        return pre

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