86 lines
2.2 KiB
Python
86 lines
2.2 KiB
Python
"""
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A permutation is an ordered arrangement of objects.
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For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4.
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If all of the permutations are listed numerically or alphabetically, we call it lexicographic order.
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The lexicographic permutations of 0, 1 and 2 are:
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012 021 102 120 201 210
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What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
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"""
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import time
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from itertools import permutations
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def timer(func):
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def wrapper(*args, **kwargs):
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start_time = time.time()
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result = func(*args, **kwargs)
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end_time = time.time()
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print(f"Execution time: {end_time - start_time:.6f} seconds")
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return result
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return wrapper
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def ordered_permutations(n: int) -> list[int]:
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digits = list(range(10))
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all_permutations = permutations(digits)
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for _ in range(n - 1):
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next(all_permutations)
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return next(all_permutations)
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@timer
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def main_iter():
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n = 1000000
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result = ordered_permutations(n)
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print(f"used iter: {''.join(map(str, result))}")
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def get_permutation(seq, k):
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"""
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返回序列 seq 的第 k 个字典序排列 (1-based)
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参数:
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seq: 可迭代对象（列表、元组、字符串等），元素需有序
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k: 第 k 个排列（从1开始计数）
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返回:
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与输入类型相同的排列结果
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"""
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n = len(seq)
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# 预计算阶乘
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factorial = [1]
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for i in range(1, n):
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factorial.append(factorial[-1] * i)
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# 将输入转为列表（复制一份避免修改原序列）
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elements = list(seq)
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k -= 1 # 转为0-based索引
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# 构建结果
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result = []
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for i in range(n, 0, -1):
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index = k // factorial[i - 1] # 当前位选择的元素索引
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result.append(elements.pop(index))
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k %= factorial[i - 1] # 更新剩余位
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# 根据输入类型返回相应格式
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if isinstance(seq, str):
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return "".join(result)
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return result
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@timer
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def main_math():
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n = 1000000
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result = get_permutation("0123456789", n)
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print(f"used math: {result}")
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if __name__ == "__main__":
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main_iter()
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main_math()
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