*/classSolution{publicListNodereverseKGroup(ListNode head,intk){// boundary judgementbooleanhasNoOrOneNode=(head ==null|| head.next ==null);if(hasNoOrOneNode || k ==1) {returnhead; }ListNodedummy=newListNode(-1); dummy.next = head;ListNodebegin=dummy;inti=0;while(head !=null) { ++i...
https://leetcode-cn.com/problems/reverse-nodes-in-k-group/description/ 给出一个链表,每 k 个节点一组进行翻转,并返回翻转后的链表。 k 是一个正整数,它的值小于或等于链表的长度。如果节点总数不是 k 的整数倍,那么将最后剩余节点保持原有顺序。 解法1 首先遍历链表,记录结点个数,当满足个数等于k时,...
原题链接:https://leetcode.com/problems/reverse-nodes-in-k-group/ 解题思路:这道题是reverse two的升级版,我的代码其实效率很低。首先用了两个标记,一个start指向当前正在reverse的group的第一个,另一个next_start指向下一组的第一个。然后要next_n标记pre要指向的,用cur标记next_n要指向的。一组一组.....
next) if has_no_or_one_node or k == 1: return head dummy = ListNode(-1) dummy.next = head begin = dummy i = 0 while head: i += 1 if i % k == 0: begin = self._reverse(begin, head.next) head = begin.next else: head = head.next return dummy.next def _reverse(self,...
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
创建一个 dummy node对链表以 k 为单位进行分组，记录每一组的起始和最后节点位置对每一组进行翻转，更换起始和最后的位置返回dummy.next.代码 (Java/Python3/javascript)Java Code class ReverseKGroupsLinkedList { public ListNode reverseKGroup(ListNode head, int k) { if (head == null || k == 1...
You may not alter the values in the list's nodes, only nodes itself may be changed. 题解: 直接用容器: classSolution{ public: ListNode*reverseKGroup(ListNode*head,intk) { if(head==NULL||head->next==NULL||k<=1) { returnhead; ...
给出一个链表,每 k 个节点一组进行翻转,并返回翻转后的链表。 k 是一个正整数,它的值小于或等于链表的长度。如果节点总数不是 k 的整数倍,那么将最后剩余节点保持原有顺序。 示例: 给定这个链表:1->2->3->4->5 当k = 2 时,应当返回: 2->1->4->3->5 ...
if(len<k) return head; ListNode *q=head; p=NULL; while(q&&n>0) { ListNode *ne=q->next; q->next=p; p=q; q=ne; n--; } if(len-k>=k) head->next=reverseKGroup(q,k); else head->next=q; return p; } }; 1.
count+=1ifcount==k:cur=self.reverseKGroup(cur,k)print("cur.val=",cur.val)whilecount:print("head.val=",head.val)tmp=head.next head.next=cur cur=head head=tmp count-=1head=curreturnhead 代码来源: 作者:powcai 链接:https://leetcode-cn.com/problems/reverse-nodes-in-k-group/solution/...