[LeetCode-interview algorithm classic-Java implementation] [019-Remove Nth Node From End of List (Remove the last N nodes of a single-chain table)], removenode

[LeetCode-interview algorithm classic-Java implementation] [019-Remove Nth Node From End of List (Remove the last N nodes of a single-chain table)], removenode
[019-Remove Nth Node From End of List (Remove the Nth Node From the last of a single-chain table )][LeetCode-interview algorithm classic-Java implementation] [directory indexes for all questions]Original question

Given a linked list, remove the nth node from the end of list and return its head.
For example,

   Given linked list: 1->2->3->4->5, and n = 2.   After removing the second node from the end, the linked list becomes 1->2->3->5.

　　Note:
Given n will always be valid.
Try to do this in one pass.

Theme

Delete the nth last node of a single-chain table. Note: The input N is valid and the operation is completed in a traversal.

Solutions

First let a pointer go to the nth node, then let a pointer point to the header node, and then two pointers go together until the previous pointer ends at the end, the next pointer is the last N + 1 node. You can delete the last N node.

Code Implementation

Linked List Node Type

public class ListNode {    int val;    ListNode next;    ListNode(int x) {        val = x;        next = null;    }}

Algorithm Implementation class

Public class Solution {public ListNode removeNthFromEnd (ListNode head, int n) {ListNode pa = head; ListNode pb = head; // locate the nth node for (int I = 0; I <n & pa! = Null; I ++) {pa = pa. next;} if (pa = null) {head = head. next; return head;} // The difference between pb and pa is n-1 node // when pa. next is null, and pb is in the n + 1 position at the bottom while (pa. next! = Null) {pa = pa. next; pb = pb. next;} pb. next = pb. next. next; return head ;}}

Evaluation Result

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Note Please refer to the following link for more information: http://blog.csdn.net/derrantcm/article/details/46997239]

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