LeetCode——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.

Note:

Given n will always be valid.

Try to do this in one pass.
原题链接：https://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/
题目：给定一个链表，从尾部删除第n个节点并返回新的链表。

思路：使用两个指针，fast 和 slow，他们的距离是n，于是fast到尾的时候，n所在的节点就是需要删除的节点。

public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode slow = head, fast = head;
if (head.next == null)
return null;
for (int i = 1; i <= n; i++)
slow = slow.next;
if (slow == null) {
head = head.next;
return head;
}
while (slow.next != null) {
slow = slow.next;
fast = fast.next;
}
fast.next = fast.next.next;
return head;
}
// Definition for singly-linked list.
public class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}