题解 | #【模板】链表#

# class Stack:
#     def __init__(self, val = None):
#         self.val = val
#         self.next = None

# phead = Stack()
# num = input()
# for i in range(int(num)):
#     line = input().split()
#     x = int(line[1])
#     cur = phead
#     if line[0][0] == 'i':
#         y = int(line[2])
#         while cur and cur.val and cur.val !=x:
#             cur = cur.next
#         if not cur:
#             cur = Stack(y)
#         elif not cur.val:
#             cur.val = y
#         elif not cur.next:
#             cur.next = Stack(y)
#         else:
#             cur2 = cur.next
#             cur.next = Stack(y)
#             cur.next.next = cur2
#     else:
#         if cur.val and cur.val == x:
#             phead = cur.next
#         else:
#             while cur.next:
#                 if cur.next.val == x:
#                     cur.next = cur.next.next
# while phead:
#     print(phead.val,end = ' ')
#     phead = phead.next

# 摆烂了，直接数组
s = []
n = int(input())
for i in range(n):
    a = input()
    if a[0] == 'i':
        a = a.split()
        x = int(a[1])
        y = int(a[2])
        if len(s) == 0:
            s = [y]
        else:
            cur = 0
            for j in range(len(s)):
                cur = j
                if s[j] == x:
                    s.insert(j,y)
                    break
            if cur == len(s) - 1:
                s.append(y)
    else:
        a = a.split()
        x = int(a[1])
        for j in range(len(s)):
            if s[j] == x:
                s.pop(j)
                break
if len(s) == 0:
    print('NULL')
else:
    for j in range(len(s)):
        print(s[j],end = ' ')

看了一下别人写的：

link = []

fori in range(int(input())):

ipt = input().split()

ifipt[0] == 'insert':

x,y = int(ipt[1]),int(ipt[2])

ifx in link:

link.insert(link.index(x),y)

else:

link.append(y)

ifipt[0] == 'delete':

x = int(ipt[1])

ifx in link:

link.remove(x)

print('NULL') ifnot link else[print(i,end=' ') fori in link]

这也太强了，学到新知识了，可以if i in Link这种语句，高效的去查找。学会了。