• ListNode *b=new ListNode(1);
• ListNode *c=new ListNode(2);
• ListNode *d=new ListNode(3);
• ListNode *e=new ListNode(4);
• ListNode *f=new ListNode(5);
• ListNode *g=new ListNode(6);
• a->next=b;
• b->next=c;
• c->next=d;
• d->next=e;
• e->next=f;
• f->next=g;
• g->next=NULL;
• Solution sol;
• ListNode *head=sol.reverseKGroup(a,4);
• while(head!=NULL){

•   cout<<head->val<<' ';
  

•   head=head->next;
  

• }
• cout<<endl;
• return 0; +} §§ --- 25. Reverse Nodes in k-Group/Readme.md §§ 1000

25. Reverse Nodes in k-Group

Difficulty for me

Medium

Complexity

time complexity = O(n)

space complexity = O(1)

runtime: beat 71.25%

Summary

reverse

每k个node反转，和206题类似，但是有个难点是找到每次反转的末尾node，其实可以用一个指针，每次指向末尾node.

Tips

1. 链表最后一段不够k个，不需要反转
2. 记得最后把尾节点的next置为空