线性搜索 Python

Question

我是 python 新手，我们被分配创建一个不使用“in”或索引的线性搜索程序。该程序编译但说我输入的每个数字都不在列表中。我也必须为二进制搜索做同样的事情，但我一次只做一件事情哈。任何帮助表示赞赏！

PS：如果不使用“索引”功能，如何显示它所在的索引？

def linearSearch(intList,target):
    found = False
    position = 0
    while position < len(intList) and not found:
        if intList[position] == target:
            found = True
        position = position + 1

    return found

linearList = [3,5,9,7,6,12,15,9,1]
numInput = input("What number are you looking for? ")
numFound = linearSearch(linearList, numInput)
if numFound:
    print("The number is in index: ")
else:
    print("The number is not in the list")

Answer 1

def linearSearch(array,k):
    
    flag = False
    position = 0
    while position < len(intList) and not flag:
       
        if array[position] == k:
            flag = True
        else:
            position += 1

    return position

array = [3,5,9,7,6,12,15,9,1]
numInput = int(input("What number are you looking for? "))
numFound = linearSearch(linearList,numInput)
if numFound:
    print("The number is in index: ")
else:
    print("The number is not in the list")

Answer 2

此方法使用列表推导，它也将考虑列表中的任何重复项。它分配一个索引/索引列表，其中键出现在列表中。阅读有关列表理解的更多信息here.

l = [1, 2, 3, 4, 4, 6]
el = 4
search = [i for i in range(len(l)) if el==l[i]]
print(search)

输出：

[3, 4]

或者，

def LinSearch(target, intList):
     search = [i for i in range(len(intList)) if target==intList[i]]
     return search

Answer 3

 n=int(input("Enter the number:"))
 list,n=[10,11,12,13,14,15],n
 if n in list:
     print("Element found at",list.index(n))
 else:
     print("Not Found")

建议：在 python 中，索引从 0 开始。要从 1 开始，格式如下：list.index(n)+1

Answer 4

def linear_search(list, key):

    for i, item in enumerate(list):
        if item == key:
            return i
    return -1

print(linear_search([4,5,2,7,1,8],7))

#If key is in the list returns its position in the list, otherwise returns -1.

Answer 5

def target(list,num):

    for x in list:
        if x == str(num):
            print("match found:"+x)
            break
    else:
         print('no match found')


list1 = ['6','4','7','9','0','2','3','1']
inp = input('Enter number to search:')
target(list1,inp)

Answer 6

线性搜索：

 // funtion which rturns true if item found inside list.
    def linearSearch(list, value):
            for i in range(len(list)):
                if i == value:
                     return True

// 调用上面的函数传递值列表和要搜索的项目

    list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
    item = 10
    print(linearSearch(list, item)) // item to search

Answer 7

def linearSearch(intList,target):
    #print (target)
    found = False
    position = 0
    while position < len(intList):
        #print(intList[position])
        if intList[position] == target:
            found = True
            break
        position = position + 1

    return found

linearList = [3,5,9,7,6,12,15,9,1]
numInput = int(input("What number are you looking for? "))
numFound = linearSearch(linearList,numInput)
if numFound:
    print("The number is in index: ")
else:
    print("The number is not in the list")

请注意类型转换...

Answer 8

1) 开始position = -1

2)return position

3) 你想在if intList[position] == target: 之前position+=1 并且当你找到元素时你想break。那么你就不需要found

linearSearch(linearList, numInput) > 0时发现了一些东西

然后，您的代码不起作用，因为列表包含整数，而 input 将始终返回一个字符串。你必须使用int(input(".."))