The penultimate node in the linked list - Double finger

The finger of the sword Offer 22. Last in the list k Nodes

The difficulty is simple 376

Enter a linked list , Output the last number in the list k Nodes . In order to conform to the habits of most people , From 1 Start counting , That is, the tail node of the list is the last 1 Nodes .

for example , A list has 6 Nodes , Start from the beginning , Their values, in turn, are 1、2、3、4、5、6. The last of the list 3 Each node has a value of 4 The node of .

Example ：

 Given a linked list : 1->2->3->4->5,  and  k = 2.

 Back to the list  4->5.

Pass times 374,318

Submit the number 466,479

Ideas

Use double pointers to solve the problem , First define the speed pointer , Let the quick pointer go first k Step , Finally, walk with the slow pointer , Until the end , Returns the full pointer

Brushing questions makes me happy

The title algorithm code is as follows

class Solution {
    
    public ListNode getKthFromEnd(ListNode head, int k) {
    
   ListNode slow =head,fast=head;
         while (fast!=null&&k>0){
    
             fast=fast.next;
             k--;
         }
         // Went together 
         while (fast!=null){
    
             slow=slow.next;
             fast=fast.next;
         }
         return slow;
    }
}