# 低效方式：列表查找
data = [i * 2 for i in range(1000000)]
print(data.index(1999998))  # O(n)

# 高效方式：字典查找
data = {i: i * 2 for i in range(1000000)}
print(data[999999])  # O(1)

# 嵌套循環(huán)（O(n2)）
def find_duplicates1(arr):
    duplicates = []
    for i in range(len(arr)):
        for j in range(i + 1, len(arr)):
            if arr[i] == arr[j]:
                duplicates.append(arr[i])
    return duplicates# 優(yōu)化方案（O(n)）
def find_duplicates2(arr):

    seen = set()
    duplicates = []
    for num in arr:
        if num in seen:
            duplicates.append(num)
        else:
            seen.add(num)
    return duplicates

# 傳統(tǒng)方式
squares = []
for x in range(10):
    squares.append(x ** 2)

# 優(yōu)化方式
squares = [x ** 2 for x in range(10)]  # 執(zhí)行速度更快

# 生成器表達(dá)式（節(jié)省內(nèi)存）
my_generator = (x ** 2 for x in range(1000000))
for value in my_generator:
    pass  # 處理數(shù)據(jù)

# 與列表推導(dǎo)式的對(duì)比
my_list = [x ** 2 for x in range(1000000)]  # 占用大量?jī)?nèi)存

# 使用內(nèi)置函數(shù)計(jì)算總和
total = sum(range(10000))  # 內(nèi)置函數(shù)優(yōu)化

# 手動(dòng)循環(huán)計(jì)算總和
total = 0
for x in range(10000):
    total += x  # 執(zhí)行速度較慢

from collections import Counter

# 使用Counter統(tǒng)計(jì)元素出現(xiàn)次數(shù)
data = [1, 2, 2, 3, 3, 3, 4, 4, 4, 4]
counter = Counter(data)  # O(n)時(shí)間復(fù)雜度
print(counter)  # 輸出: Counter({4: 4, 3: 3, 2: 2, 1: 1})

from functools import lru_cache

@lru_cache(maxsize=None)  # 緩存斐波那契數(shù)列計(jì)算結(jié)果
def fibonacci(n):
    if n < 2:
        return n
    return fibonacci(n - 1) + fibonacci(n - 2)

# 調(diào)用緩存版本的斐波那契函數(shù)
print(fibonacci(100))  # 未緩存時(shí)，O(2^n)復(fù)雜度會(huì)導(dǎo)致超時(shí)

import numpy as np

# 純Python循環(huán)
result = [x * 2 for x in range(1000000)]  # 執(zhí)行速度較慢

# NumPy向量化操作
arr = np.arange(1000000)
result = arr * 2  # 利用底層C代碼加速

# 低效方式：逐個(gè)拼接
result = ""
for s in ["a", "b", "c", ...]:  # 1000個(gè)字符串
    result += s  # 每次拼接生成新字符串

# 高效方式：使用join
result = "".join(["a", "b", "c", ...])  # 單次分配內(nèi)存

import re

# 預(yù)編譯正則表達(dá)式
pattern = re.compile(r"\d+")

# 匹配1000個(gè)文本
for text in texts:
    match = pattern.findall(text)  # 避免重復(fù)編譯

# 低效方式：嵌套循環(huán)（O(n2)）
def intersection1(arr1, arr2):
    result = []
    for x in arr1:
        for y in arr2:
            if x == y:
                result.append(x)
    return result

# 優(yōu)化方式：集合查找（O(n)）
def intersection2(arr1, arr2):
    set1 = set(arr1)
    return [x for x in arr2 if x in set1]

import timeit

def test1():
    return [x ** 2 for x in range(1000)]

def test2():
    return list(map(lambda x: x ** 2, range(1000)))

# 比較兩種方法的執(zhí)行時(shí)間
print(timeit.timeit(test1, number=10000))
print(timeit.timeit(test2, number=10000))