LeetCode-19 Remove Nth Node From End of List(移除尾部第N个节点)

LeetCode-19 Remove Nth Node From End of List(移除尾部第N个节点)
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.

思路：设两个节点，一个节点先前进N个位置，然后另一个节点在开始出发，当第一个节点到底终点时，第二个节点位置即为倒数N位。
我的代码：

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;
		}
    }
}

后面判断部分是根据提交出错的情况加的，佩服那些可以一个代码不用那么多case就解决的。。

Runtime: 303 ms



别人的代码（比我简洁多了，运行速度也快一点）：

/**
 * 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;
    }
}