2、反转链表II:https://leetcode-cn.com/problems/reverse-linked-list-ii/ 参考# https://leetcode-cn.com/problems/reverse-nodes-in-k-group/solution/tu-jie-kge-yi-zu-fan-zhuan-lian-biao-by-user7208t/
25. Reverse Nodes in k-Group 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 node...
end] start = reverse(start, end.next); end = start.next; } else { end = end.next; } } return dummy.next; } /** * reverse linked list from range (start, end), return last node. * for example: ...
*/publicListNodereverseKGroup(ListNodehead,intk){if(head==null||head.next==null){returnhead;}ListNodetail=head;for(inti=0;i<k;i++){// 如果剩余数量小于 k,则不需要反转。if(tail==null){returnhead;}tail=tail.next;}// 反转前 k 个元素ListNodenewHead=reverse(head,tail);// 下一轮的开始...
# Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = NoneclassSolution:defreverseKGroup(self, head, k):""" :type head: ListNode :type k: int :rtype: ListNode """has_no_or_one_node = (notheadornothead.next)ifhas_no_or...
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 Solution Iterative 蛮有意思的题,最初觉得就是reverse一整条list的变种,后来发现不太一样,思路被打开了。常规reverse list的写法是用双指针法,边移...
{// 记录下一个子链表的头ListNode nextGroup=pointer.next;// 反转当前子链表,reverse 函数返回反转后子链表的头ListNode reversedList=reverse(lastGroup.next,nextGroup);// lastGroup 是上一个子链表的尾,其 next 指向当前反转子链表的头// 但是因为当前链表已经被反转,所以它指向的是反转后的链表的尾pointer...
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def reverseKGroup(self, head: ListNode, k: int) -> ListNode: newhead= ListNode(0) newhead.next = head ...
Reverse Linked List Reverse a singly linked list. Note Create tail = null; Head loop through the list: Store head.next, head points to tail, tail becomes head, head goes to stored head.next; Return tail. Solution public class Solution { ...
# Definitionforsingly-linked list.#classListNode:# def__init__(self,x):# self.val=x # self.next=NoneclassSolution:defreverseKGroup(self,head:ListNode,k:int)->ListNode:newhead=ListNode(0)newhead.next=head pre=newhead tail=newheadwhileTrue:count=kwhilecount and tail:count-=1tail=tail.nex...