LeetCode-25- Reverse Nodes in k-Group

痛定思痛。 2021-12-15 14:01 244阅读 0赞

算法描述：

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 then left-out nodes in the end should remain as it is.

Example:

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

Note:

  • Only constant extra memory is allowed.
  • You may not alter the values in the list’s nodes, only nodes itself may be changed.

解题思路：细节实现题。

  1. ListNode* reverseKGroup(ListNode* head, int k) {
  2.     if(head == nullptr)  return head;
  3.     int count =0;
  4.     ListNode* cur = head;
  5.     while(cur!=nullptr && count < k){
  6.         cur=cur->next;
  7.         count++;
  8.     }
  9.     if(count==k){
  10.         cur = reverseKGroup(cur,k);
  11.         while(count--){
  12.             ListNode* temp = head->next;
  13.             head->next=cur;
  14.             cur=head;
  15.             head=temp;
  16.         }
  17.         head=cur;
  18.     }
  19.     return head;
  20. }

转载于:https://www.cnblogs.com/nobodywang/p/10362104.html

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