二进制搜索python 3.5

我试图用Python3.5编写二进制搜索，但它不起作用，我不知道为什么

def binarySearch(alist, value):

    first = 0
    last = len(alist)-1
    midpoint = (last//2)
    while binarySearch:
        if value == alist[midpoint]:
            return True and print ("found")
        else:
            if value < midpoint:
                last = midpoint-1
            else:
                if value > midpoint:
                    first = midpoint+1    
binarySearch([1,2,3,4,5,6,7,8],3)

def二进制搜索（列表，值）：
第一个=0
last=len（alist）-1
中点=（最后//2）
二进制搜索时：
如果值==列表[中点]：
返回True并打印（“找到”）
其他：
如果值<中点：
最后=中点-1
其他：
如果值>中点：
第一个=中点+1
二进制搜索（[1,2,3,4,5,6,7,8]，3）

若我将值设置为4，它将显示find，若我设置了其他任何内容，则不会发生任何事情，并且它会在运行时被卡住，什么也不做

谢谢你的帮助

当binarySearch时，您的循环条件是错误的

您只需更改中点的值一次，而应该在每次循环迭代中更改它

将值与索引（中点）进行比较，并应与 列表值（列表[中点]）

这是错误的：

返回True并打印（“找到”）

它将始终不返回任何值

---

User1915011抢先回答了我的问题。根据他的回答和@wim的评论，我对您的

binarySearch

方法做了以下更改

将循环更改为使用

found

变量

向循环内的

中点添加了一个附加赋值

---

通过添加

first二进制转换器来确保循环终止

num = int(input('please enter your number: '))  
list = []  
for i in (128, 64, 32, 16, 8, 4, 2, 1):  
    if num >= i:  
        list.append(1)  
        num = num-i  
    else:  
        list.append(0)
print(list)

以下是其工作原理的详细说明：

    def binarySearch(array, i):

    ## Binary search is the algorithm which is used to search an element in a sorted array

    ## The time complexity of the binary search is O(log n)
    ## Which means that in an array of length(2^(n-1)) elements we need to look at only n elements
    ## That is why we say that binary search algorithm runs in logarithmic time, which is much faster than linear time

    start = 0
    last = len(array)-1
    result = False

    count = 0 ## to debug

    print("\n******************************************************************\n")
    while(start <= last and not result):

        ## Debugger Begin
        mid = 0
        print("Loop number: ", count)
        print("Start element: ", array[start], " Position of Start Element: ", start)
        print("Last element: ", array[last], "  Position of Last Element: ", last)
        ## Debugger End

        mid = (start + last)//2 ## '//' indicates the floor division(ignores the value after the period) 
        if(array[mid] == i):
            print("***Mid***")
            result = True;
        else:
            if(i < array[mid]):
                print("***The value of the item:",i," we are searching for is LESS than the current middle element***")
                last = mid - 1
            else:
                print("***The value of the item:",i," we are searching for is GREATER than the current middle element***")
                start = mid + 1

        ## Debugger
        count = count+1
        print("Mid element: ", array[mid], "   Position of Mid Element: ", mid, "\n")
        ## Debugger

    print("******************************************************************")
    if(result == True):
        print("\nThe element:",i ,"is in the array")
    else:
        print("\nItem is not in the array")

    return result

## Array you want to search
array = [9, 11, 12, 21, 23, 34, 45, 49, 65, 98]

## Item you want to search in the array
i = 21

print("Searching the element: ",i , "\nIn the Array: ", array)
print("Length of the array is: ", len(array))

## Binary Search
binarySearch(array, i)

def二进制搜索（数组，i）：
##二进制搜索是用于在排序数组中搜索元素的算法
##二进制搜索的时间复杂度为O（logn）
##这意味着在长度（2^（n-1））元素的数组中，我们只需要查看n个元素
##这就是为什么我们说二进制搜索算法在对数时间内运行，这比线性时间快得多
开始=0
last=len（数组）-1
结果=错误
计数=0##以进行调试
打印（“\n******************************************************************************************************************************\n”）
虽然（开始时，您从不在中重新指定
中点
）loop@wim啊，好吧，我会试试的，我知道这可能是为了上课，但一般来说，不要写你自己的二进制搜索。Python已经有了确切的目的。是的，我理解，但它会在我的考试中出现，他们希望我能写出来。非常感谢老兄appreciated@Jaco因为如果在循环中指定中点，则无需预先指定。请注意，它不会使用last=len（alist）-1识别最后一个元素，它会将其识别为最后一个元素之前的元素。因此，在给定的示例中，它会认为7是最后一个元素，而不是8？谢谢。
binarySearch（[1,2,3,4,5,6,7,8]）
似乎有效？在Python中，索引从0开始，因此
a[len（a）-1]
如果a=[1,2,3,4,5,6,7,8]则返回8
    def binarySearch(array, i):

    ## Binary search is the algorithm which is used to search an element in a sorted array

    ## The time complexity of the binary search is O(log n)
    ## Which means that in an array of length(2^(n-1)) elements we need to look at only n elements
    ## That is why we say that binary search algorithm runs in logarithmic time, which is much faster than linear time

    start = 0
    last = len(array)-1
    result = False

    count = 0 ## to debug

    print("\n******************************************************************\n")
    while(start <= last and not result):

        ## Debugger Begin
        mid = 0
        print("Loop number: ", count)
        print("Start element: ", array[start], " Position of Start Element: ", start)
        print("Last element: ", array[last], "  Position of Last Element: ", last)
        ## Debugger End

        mid = (start + last)//2 ## '//' indicates the floor division(ignores the value after the period) 
        if(array[mid] == i):
            print("***Mid***")
            result = True;
        else:
            if(i < array[mid]):
                print("***The value of the item:",i," we are searching for is LESS than the current middle element***")
                last = mid - 1
            else:
                print("***The value of the item:",i," we are searching for is GREATER than the current middle element***")
                start = mid + 1

        ## Debugger
        count = count+1
        print("Mid element: ", array[mid], "   Position of Mid Element: ", mid, "\n")
        ## Debugger

    print("******************************************************************")
    if(result == True):
        print("\nThe element:",i ,"is in the array")
    else:
        print("\nItem is not in the array")

    return result

## Array you want to search
array = [9, 11, 12, 21, 23, 34, 45, 49, 65, 98]

## Item you want to search in the array
i = 21

print("Searching the element: ",i , "\nIn the Array: ", array)
print("Length of the array is: ", len(array))

## Binary Search
binarySearch(array, i)