LeetCode-----19. Remove Nth Node From End of List

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.(输出)

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

Input:[1]1

Output:[1]

Expected:[]

Input:[1,2]1

Output:[]

Expected:[1]

// pb与pa相差n-1个结点// 当pa.next为null，pb在倒数第n+1个位置

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode removeNthFromEnd(ListNode head, int n) {        ListNode pa = head;        ListNode pb = head;        // 找到第n个结点        for (int i = 0; i < n && pa != null; i++) {            pa = pa.next;        }        if (pa == null) {            head = head.next;            return head;        }        // pb与pa相差n-1个结点        // 当pa.next为null，pb在倒数第n+1个位置        while (pa.next != null) {            pa = pa.next;            pb = pb.next;        }        pb.next = pb.next.next;//删除caozuo        return head;    }}

public class Solution {    public ListNode removeNthFromEnd(ListNode head, int n) {        if (n <= 0) {            return null;        }                ListNode dummy = new ListNode(0);        dummy.next = head;                ListNode preDelete = dummy;        for (int i = 0; i < n; i++) {            if (head == null) {                return null;            }            head = head.next;        }        while (head != null) {            head = head.next;            preDelete = preDelete.next;        }        preDelete.next = preDelete.next.next;        return dummy.next;    }}