How does leetCode find the last-to-last node in the linked list
Editor to share with you leetCode how to find the last-to-last node in the linked list, I believe most people do not know much about it, so share this article for your reference, I hope you can learn a lot after reading this article, let's go to know it!
First, the last but one node 1 in the linked list, a brief description of the problem
Enter a linked list and output the last but one node in the linked list. In order to meet the habits of most people, this topic starts counting from 1, that is, the end node of the linked list is the penultimate node. For example, a linked list has six nodes, starting from the top node, and their values are 1, 2, 3, 4, 5, and 6. The penultimate node of this linked list is the node with a value of 4.
2. Example description example:
Given a linked list: 1-> 2-> 3-> 4-> 5, and k = 2.
Return to linked list 4-> 5.
3, the train of thought of solving the problem
The problem can be solved by using fast and slow pointers.
4, problem solving procedure
Import java.util.ArrayList
Import java.util.LinkedList
Import java.util.List
Public class GetKthFromEndTest {
Public static void main (String [] args) {
ListNode L1 = new ListNode (1)
ListNode L2 = new ListNode (2)
ListNode L3 = new ListNode (3)
ListNode L4 = new ListNode (4)
ListNode L5 = new ListNode (5)
L1.next = L2
L2.next = L3
L3.next = L4
L4.next = L5
Int k = 2
ListNode listNode = getKthFromEnd (L1, k)
System.out.println ("listNode =" + listNode)
}
Public static ListNode getKthFromEnd (ListNode head, int k) {
If (head = = null | | head.next = = null) {
Return head
}
ListNode slow = head
ListNode fast = head
For (int I = 0; I < k; iTunes +) {
Fast = fast.next
}
While (fast! = null) {
Slow = slow.next
Fast = fast.next
}
Return slow
}
}
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