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70 lines
1.5 KiB
70 lines
1.5 KiB
"""
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File: recursion.py
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Created Time: 2023-08-24
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Author: Krahets (krahets@163.com)
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"""
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def recur(n: int) -> int:
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"""递归"""
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# 终止条件
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if n == 1:
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return 1
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# 递：递归调用
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res = recur(n - 1)
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# 归：返回结果
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return n + res
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def for_loop_recur(n: int) -> int:
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"""使用迭代模拟递归"""
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# 使用一个显式的栈来模拟系统调用栈
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stack = []
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res = 0
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# 递：递归调用
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for i in range(n, 0, -1):
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# 通过“入栈操作”模拟“递”
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stack.append(i)
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# 归：返回结果
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while stack:
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# 通过“出栈操作”模拟“归”
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res += stack.pop()
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# res = 1+2+3+...+n
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return res
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def tail_recur(n, res):
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"""尾递归"""
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# 终止条件
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if n == 0:
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return res
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# 尾递归调用
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return tail_recur(n - 1, res + n)
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def fib(n: int) -> int:
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"""斐波那契数列：递归"""
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# 终止条件 f(1) = 0, f(2) = 1
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if n == 1 or n == 2:
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return n - 1
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# 递归调用 f(n) = f(n-1) + f(n-2)
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res = fib(n - 1) + fib(n - 2)
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# 返回结果 f(n)
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return res
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"""Driver Code"""
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if __name__ == "__main__":
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n = 5
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res = recur(n)
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print(f"\n递归函数的求和结果 res = {res}")
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res = for_loop_recur(n)
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print(f"\n使用迭代模拟递归求和结果 res = {res}")
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res = tail_recur(n, 0)
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print(f"\n尾递归函数的求和结果 res = {res}")
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res = fib(n)
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print(f"\n斐波那契数列的第 {n} 项为 {res}")
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