High-quality code-the penultimate K-node in a linked list

Title Description:

Enter a list to output the last K nodes in the linked list.

Ideas:

First check the legitimacy of the parameters, Head==null or the number of nodes less than K directly return NULL.

Let head advance k-1 step, ans points to the head node, then head forward one step, ans also step forward. When the head reaches the last node, ans points to the reciprocal K-node. Time complexity O (n).

Solve:

1 /*2 Public class ListNode {3 int val;4 ListNode next = null;5 6 listnode (int val) {7 this.val = val;8     }9 }*/Ten  Public classSolution { One      PublicListNode Findkthtotail (ListNode head,intk) { A         -ListNode ans =head; -          the         if(Head = =NULL|| K < 1) { -             return NULL; -         } -         inti =K; +         //head forward first k-1 node -          while(i > 1 && head.next! =NULL) { +Head =Head.next; Ai--; at         } -         //Insufficient number of nodes -         if(I! = 1) { -             return NULL; -         } -          in          while(Head.next! =NULL) { -Head =Head.next; toAns =Ans.next; +         } -         the         returnans; *          $     }Panax Notoginseng}

High-quality code-the penultimate K-node in a linked list