1 """
2 Given a linked list, remove the n-th node from the end of list and return its head.
3 Example:
4 Given linked list: 1->2->3->4->5, and n = 2.
5 After removing the second node from the end, the linked list becomes 1->2->3->5.
6 """
7 """
8 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
9 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
10 """
11 """
12 Time Limit Exceeded
13 Last executed input
14 [1]
15 1
16 """
17 class ListNode:
18 def __init__(self, x):
19 self.val = x
20 self.next = None
21
22 class Solution1:
23 def removeNthFromEnd(self, head, n):
24 total = 0
25 p = head
26 q = head
27 while p:
28 total += 1
29 p = p.next
30 step = total - n -1
31 while step:
32 if q:
33 q = q.next
34 step -= 1
35 q.next = q.next.next
36 return head
37
38 class ListNode:
39 def __init__(self, x):
40 self.val = x
41 self.next = None
42
43 class Solution2:
44 def removeNthFromEnd(self, head, n):
45 pre = head
46 cur = head
47 # for _ in range(n):
48 # cur = cur.next
49 # bug 注意for 和 while循环的使用
50 while n: #利用间距 可以省去一次循环
51 cur = cur.next
52 n -= 1
53 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
54 return head.next
55 while cur.next:
56 cur = cur.next
57 pre = pre.next
58 pre.next = pre.next.next
59 return head
