为什么我的二进制搜索比线性搜索慢？

问题描述

我试图编写二进制搜索和线性搜索的代码，看到二进制搜索有时甚至比线性搜索慢两倍，这让我感到震惊。请帮帮我。以下是我的代码。

二进制搜索代码：

def binary_search(array, target, n=0):
    l = len(array)-1
    i = l//2
    try:
        ai = array[i]
    except:
        return False
    if ai == target:
        n += i
        return (True, n)

    elif target >= ai:
        array = array[i+1:l+1]
        n += i + 1
        return binary_search(array, target, n)

    elif target <= ai:
        array = array[0: i]
        return binary_search(array, target, n)

线性搜索代码

def linear_search(array, target):
    for i, num in enumerate(array):
        if num == target:
            return True, i

    return False

测试用例代码：

import random 
import time
n = 10000000
num = sorted([random.randint(0, n) for x in range(n)])

start = time.time()
print(linear_search(num, 1000000))
print(f'Linear Search: {time.time() - start}')

start_new = time.time()
print(binary_search(num, 1000000))
print(f'Binary Search: {time.time() - start_new}')

---

解决方案

正如@khelwood所说，没有切片，您的代码将会更快。

def binary_search_no_slice(array, target, low, high):
    if low > high:
        return False
    mid = (low + high) // 2
    if array[mid] == target:
        return True
    elif array[mid] > target:
        return binary_search_no_slice(array, target, low, mid - 1)
    else:
        return binary_search_no_slice(array, target, mid + 1, high)

已将以下内容添加到您的测试代码中。

start_new2 = time.time()
print(binary_search_no_slice(num, 1000000, 0, len(num) - 1))
print(f'Binary Search no slice: {time.time() - start_new2}')

这是在我的机器(MacOS Catalina，2.8 GHz Corei7，8 GB内存)上的结果

False
Linear Search: 2.172485113143921
False
Binary Search: 0.56640625
False
Binary Search no slice: 2.8133392333984375e-05