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