首页 > 解决方案 > 在我的二进制搜索算法中,Python 在列表中找不到 0 个索引成员

问题描述

如何修复代码 - 算法找到所有情况下的搜索号码,除非您尝试查找 0 索引号码或在我的情况下 search_num = 1。

list1 = [1, 3, 4, 5, 19, 21, 52, 100, 152]
search_num = 1

def binarySearch(array, number):

    lower_b_index = 0
    upper_b_index = len(array) - 1

    while lower_b_index <= upper_b_index:

        mid_point = (lower_b_index + upper_b_index) // 2

        if array[mid_point] == number:
            return mid_point
        elif array[lower_b_index] == number:
            return lower_b_index
        else:
            if array[mid_point] < number:
                lower_b_index = mid_point + 1
            else: 
                upper_b_index = mid_point - 1 

    return False

a1 = binarySearch(list1, search_num)

if a1:
    print("The searched position: ", a1 + 1)
else:
    print("Not Found!")

标签: python-3.xbinary-search

解决方案


您的问题不在您的binarySearch算法中,而是您的if声明。

您的算法返回a1 = 0正确的并且您的检查if 0和 0 是布尔值 False

一个简单的解决方法是return None检查if a1 is not None

list1 = [1, 3, 4, 5, 19, 21, 52, 100, 152]
search_num = 1

def binarySearch(array, number):

    lower_b_index = 0
    upper_b_index = len(array) - 1

    while lower_b_index <= upper_b_index:

        mid_point = (lower_b_index + upper_b_index) // 2

        if array[mid_point] == number:
            return mid_point
        elif array[lower_b_index] == number:
            return lower_b_index
        else:
            if array[mid_point] < number:
                lower_b_index = mid_point + 1
            else: 
                upper_b_index = mid_point - 1 

    return None  # Change here

a1 = binarySearch(list1, search_num)

if a1 is not None:  # Change here
    print("The searched position: ", a1 + 1)
else:
    print("Not Found!")

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