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refactor: Follow the PEP 585 Typing standard (#439)
* Follow the PEP 585 Typing standard * Update list.py
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@@ -4,60 +4,55 @@ Created Time: 2022-11-25
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Author: Krahets (krahets@163.com)
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"""
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import sys, os.path as osp
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sys.path.append(osp.dirname(osp.dirname(osp.abspath(__file__))))
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from modules import *
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""" Driver Code """
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if __name__ == "__main__":
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""" 初始化列表 """
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list: List[int] = [1, 3, 2, 5, 4]
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print("列表 list =", list)
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arr: list[int] = [1, 3, 2, 5, 4]
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print("列表 arr =", arr)
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""" 访问元素 """
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num: int = list[1]
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num: int = arr[1]
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print("访问索引 1 处的元素,得到 num =", num)
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""" 更新元素 """
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list[1] = 0
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print("将索引 1 处的元素更新为 0 ,得到 list =", list)
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arr[1] = 0
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print("将索引 1 处的元素更新为 0 ,得到 arr =", arr)
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""" 清空列表 """
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list.clear()
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print("清空列表后 list =", list)
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arr.clear()
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print("清空列表后 arr =", arr)
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""" 尾部添加元素 """
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list.append(1)
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list.append(3)
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list.append(2)
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list.append(5)
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list.append(4)
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print("添加元素后 list =", list)
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arr.append(1)
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arr.append(3)
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arr.append(2)
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arr.append(5)
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arr.append(4)
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print("添加元素后 arr =", arr)
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""" 中间插入元素 """
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list.insert(3, 6)
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print("在索引 3 处插入数字 6 ,得到 list =", list)
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arr.insert(3, 6)
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print("在索引 3 处插入数字 6 ,得到 arr =", arr)
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""" 删除元素 """
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list.pop(3)
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print("删除索引 3 处的元素,得到 list =", list)
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arr.pop(3)
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print("删除索引 3 处的元素,得到 arr =", arr)
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""" 通过索引遍历列表 """
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count: int = 0
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for i in range(len(list)):
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for i in range(len(arr)):
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count += 1
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""" 直接遍历列表元素 """
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count: int = 0
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for n in list:
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for n in arr:
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count += 1
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""" 拼接两个列表 """
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list1: List[int] = [6, 8, 7, 10, 9]
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list += list1
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print("将列表 list1 拼接到 list 之后,得到 list =", list)
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arr1: list[int] = [6, 8, 7, 10, 9]
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arr += arr1
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print("将列表 arr1 拼接到 arr 之后,得到 arr =", arr)
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""" 排序列表 """
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list.sort()
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print("排序列表后 list =", list)
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arr.sort()
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print("排序列表后 arr =", arr)
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