python如何实现数组反转

arr = [1,2,3]
arr.reverse()
print(arr)

arr = [1,2,3]
arr=arr[::-1]
print(arr)

# 增加 append
A = [1, 2, 3, 4]
A.append(5)  # 在数组末尾追加元素
print(A)

# 删除 remove pop del
A = [1, 2, 3, 4, 5, 1]
A.remove(1)  # 用于移除列表中某个值的第一个匹配项
# print(A.remove(1))  None
print(A)

# pop
A = [1, 2, 3, 4, 5]
A.pop()
print(A.pop())  # 4
print(A)  # [1, 2, 3]

print(A.pop(1))  # 2
print(A)  # [1, 3]

# del
A = [1, 2, 3, 2, 4, 5]
del A[3]
print(A)  # [1, 2, 3, 4, 5]

# 插入 insert
A = [1, 2, 3, 4, 5]
A.insert(3, 9)
print(A)  # [1, 2, 3, 9, 4, 5]

# 查找
A = [1, 2, 3, 4, 5]
if 5 in A:
    print('存在')

A = [5, 1, 2, 3, 4, 5]

print(A.index(5))  # 查找元素第一次出现的索引

# 修改
A = [1, 2, 3, 4, 5]
A[1] = 9
print(A)  # [1, 9, 3, 4, 5]

# 反转
A = [1, 2, 3, 4, 5]
A.reverse()
print(A)  # [5, 4, 3, 2, 1]

# 排序
A = [1, 2, 3, 4, 5, 0]
# print(A.sort()) # None
A.sort()  # 升序
print(A)  # [0, 1, 2, 3, 4, 5]

A.sort(reverse=True)  # 降序
print(A)  # [5, 4, 3, 2, 1, 0]

A2 = [1, 2, 3, 4, 5, 0]
sorted(A2)  # 不对原数组进行操作，开辟额外空间
print(A2)  # [1, 2, 3, 4, 5, 0]

print(sorted(A2))  # [0, 1, 2, 3, 4, 5]

# 清空 clear
A = [1, 2, 3, 4, 5]
A.clear()
print(A)  # []

# 截取
a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
print(a[1:3:1])  # [2, 3]
print(a[:7:2])  # [1, 3, 5, 7]

print(a[::-1])  # [9, 8, 7, 6, 5, 4, 3, 2, 1]
print(a[:-1])  # [1, 2, 3, 4, 5, 6, 7, 8]
print(a[-1])  # 9