同时利用一个dummy来作为pre,再不断更新pre和head, 最后返回dummy.next. #Definition for singly-linked list.#class ListNode:#def __init__(self, val=0, next=None):#self.val = val#self.next = nextclassSolution:defreverseKGroup(self, head: ListNode, k: int) ->ListNode:ifk == 1:returnhead ...
Leetcode 25:Reverse Nodes in k-Group 1.题目描述 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...
* Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *reverseKGroup(ListNode *head, int k) { if(head==NULL ||head->next==NULL||k<=1) return head; int ...
Given a linked list, reverse the nodes of a linked listkat a time and return its modified list. kis 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 ofkthen left-out nodes in the end should remain as it is....
如果这道题你按照 92.reverse-linked-list-ii 提到的 p1, p2, p3, p4（四点法） 的思路来思考的话会很清晰。代码如下（Python）：class Solution: def reverseKGroup(self, head: ListNode, k: int) -> ListNode: if head is None or k < 2: return head dummy = ListNode(0) dummy...
publicclassSolution{publicListNodereverseBetween(ListNode head,int m,int n){if(m==n||head==null||head.next==null){returnhead;}ListNode pre=newListNode(0);pre.next=head;ListNode Mpre=pre;ListNode NodeM=head;ListNode NodeN=head;int mNum=1;int nNum=1;while(mNum<m&&NodeM!=null){Mpre=...
Can you solve this real interview question? Reverse Nodes in k-Group - Given the head of a linked list, reverse the nodes of the list k at a time, and return the modified list. k is a positive integer and is less than or equal to the length of the linke
next; } if (i == k) { curr = reverseKGroup(curr, k); while (i > 0) { ListNode tmp = head.next; head.next = curr; curr = head; head = tmp; --i; } head = curr; } return head; } } Python 实现 # Definition for singly-linked list. # class ListNode: # def __init__...
next p.next = prev prev = p p = nex return tail, head def reverseKGroup(self, head: ListNode, k: int) -> ListNode: hair = ListNode(0) hair.next = head pre = hair while head: tail = pre # 查看剩余部分长度是否大于等于 k for i in range(k): tail = tail.next if not tail:...
0024 Swap Nodes in Pairs Go 60.3% Medium 0025 Reverse Nodes in k-Group Go 53.4% Hard 0026 Remove Duplicates from Sorted Array Go 50.3% Easy 0027 Remove Element Go 52.0% Easy 0028 Find the Index of the First Occurrence in a String Go 37.4% Medium 0029 Divide Two Integers Go 17.4...