Sidney Zhang
accc74ff43
📝 docs(solutions)：添加排列问题README文档说明欧拉数算法 ♻️ refactor(solutions)：重构排列问题代码，添加数学计算方法 ✨ feat(solutions)：新增斐波那契问题解决方案，支持大数计算和Binet公式 🔧 chore(solutions)：为两个解决方案添加性能计时器装饰器
86 lines
2.2 KiB
Python
86 lines
2.2 KiB
Python
"""
|
||
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()
|