LeetCode-19 remove Nth node from End of List (remove tail nth node)

LeetCode-19 Remove Nth node from End of List (remove tail nth node)

Given A linked list, remove the nth node from the end of the 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 would always be valid.
Try to do the in one pass.

Idea: Set two nodes, one node advance n position, then the other node starts, when the first node in the end, the second node position is the reciprocal n-bit.

My Code:

public class Solution {public
    ListNode removenthfromend (listnode head, int n) {
		ListNode flag = new ListNode (head . val);
		Flag.next = Head.next;
		ListNode fir = new ListNode (0);
		Fir.next = head;
		ListNode sec = new ListNode (0);
		Sec.next = head;
		for (int i = 0; i < n; i++) {
			fir = Fir.next;
		}
		while (fir.next!=null) {
			fir = Fir.next;
			SEC = Sec.next;
		}
		if (sec.next! = null) {
			sec.next = sec.next.next;
		}
		if (Sec.next = = Flag.next) {
			return head.next;
		} else {
			return head;}}
}

The later part of the judgment is based on the situation of the submission error, admire those can be a code without so many cases to solve.

Runtime: 303 ms

Other people's code (more concise than I, run faster):

/**
 * 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) {
        if (head = = NULL ) return null;
        Final ListNode _head = new ListNode (0);
        _head.next = head;

        ListNode fast = _head;
        ListNode slow = _head;
        
        for (int i = 0; i< n; i++) fast = Fast.next;
        
        while (fast! = NULL && Fast.next! = null) {
            fast = Fast.next;
            slow = Slow.next;
        }
        
        Slow.next = Slow.next.next;
        
        return _head.next;
    }
}