leetcode19 Remove Nth Node From End of List

"""
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.
"""
"""
看到这道题想法是用遍历两遍链表，第一次找出删除点，第二次删除
其实可以注意 间距这个条件，用两个指针，一次遍历就能完成
"""
"""
Time Limit Exceeded
Last executed input
[1]
1
"""
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution1:
    def removeNthFromEnd(self, head, n):
        total = 0
        p = head
        q = head
        while p:
            total += 1
            p =  p.next
        step = total - n -1
        while step:
            if q:
                q = q.next
                step -= 1
        q.next = q.next.next
        return head

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution2:
    def removeNthFromEnd(self, head, n):
        pre = head
        cur = head
        # for _ in range(n):
        #     cur = cur.next
        # bug  注意for 和 while循环的使用
        while n:            #利用间距 可以省去一次循环
            cur = cur.next
            n -= 1
        if cur == None:  # 需要删除的节点为第一个节点时，即返回第二个节点为头节点的链表
            return head.next
        while cur.next:
            cur = cur.next
            pre = pre.next
        pre.next = pre.next.next
        return head