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Add typing annotations to Python codes. (#411)
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@ -8,52 +8,52 @@ 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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def function():
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def function() -> int:
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""" 函数 """
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# do something
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return 0
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def constant(n):
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def constant(n: int) -> None:
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""" 常数阶 """
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# 常量、变量、对象占用 O(1) 空间
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a = 0
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nums = [0] * 10000
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a: int = 0
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nums: List[int] = [0] * 10000
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node = ListNode(0)
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# 循环中的变量占用 O(1) 空间
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for _ in range(n):
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c = 0
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c: int = 0
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# 循环中的函数占用 O(1) 空间
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for _ in range(n):
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function()
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def linear(n):
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def linear(n: int) -> None:
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""" 线性阶 """
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# 长度为 n 的列表占用 O(n) 空间
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nums = [0] * n
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nums: List[int] = [0] * n
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# 长度为 n 的哈希表占用 O(n) 空间
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mapp = {}
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mapp: Dict = {}
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for i in range(n):
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mapp[i] = str(i)
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def linear_recur(n):
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def linear_recur(n: int) -> None:
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""" 线性阶(递归实现) """
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print("递归 n =", n)
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if n == 1: return
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linear_recur(n - 1)
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def quadratic(n):
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def quadratic(n: int) -> None:
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""" 平方阶 """
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# 二维列表占用 O(n^2) 空间
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num_matrix = [[0] * n for _ in range(n)]
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num_matrix: List[List[int]] = [[0] * n for _ in range(n)]
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def quadratic_recur(n):
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def quadratic_recur(n: int) -> int:
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""" 平方阶(递归实现) """
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if n <= 0: return 0
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# 数组 nums 长度为 n, n-1, ..., 2, 1
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nums = [0] * n
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nums: List[int] = [0] * n
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return quadratic_recur(n - 1)
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def build_tree(n):
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def build_tree(n: int) -> Optional[TreeNode]:
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""" 指数阶(建立满二叉树) """
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if n == 0: return None
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root = TreeNode(0)
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@ -74,5 +74,5 @@ if __name__ == "__main__":
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quadratic(n)
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quadratic_recur(n)
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# 指数阶
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root = build_tree(n)
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root: Optional[TreeNode] = build_tree(n)
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print_tree(root)
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