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.
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
- Only constant extra memory is allowed.
- You may not alter the values in the list’s nodes, only nodes itself may be changed.
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def reverseKGroup(self, head, k): """ :type head: ListNode :type k: int :rtype: ListNode """ if head is None: return head res = None prev = None p = head while True: one_node = p i = 1 k_node = p while i < k and k_node is not None: k_node = k_node.next i += 1 if k_node is None: break kone_node = k_node.next curr_node, next_node = one_node, one_node.next for i in range(1, k): node = next_node.next next_node.next = curr_node curr_node, next_node = next_node, node one_node.next = kone_node if prev is not None: prev.next = k_node if res is None: res = k_node prev = one_node p = kone_node return res if res is not None else head