Given a linked list, remove the n-th node from the end of list and return its head.

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.
Follow up:
Could you do this in one pass?

思路

有删除第一个节点的可能性，所以先设置一个头结点。
再根据n的值设置双指针，n为双指针之间的步数差。

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        target = ListNode(0)
        pre = target
        target.next = head
        cur = head
        while cur.next:
            if n > 1:
                n -= 1
            else:
                target = target.next
            cur = cur.next
        target.next = target.next.next
        return pre.next