必须输出句子中所有数字的线性搜索代码

Question

我的线性搜索必须输出所有一个字符的索引。因此，如果句子是Hello，那么它将打印L在索引2和3中。

我不知道该怎么做。

def linear_search(intList,target):
    found = False
    count = 0
    #starting while loop
    while count < len(intList):
        if intList[count] == target:
            print("The character is", count, "on the index.")
            found = True
            #break while loop
            break
        else:
            count += 1

    if not found:
        print("The character is -1.")

    return count

#inputting sentence and character
sentence = input('Enter a sentence: ')
character = input('Enter a character: ')
character_found = linear_search(sentence,character)

Answer 1

它只打印第一个字符，因为你突破了while循环。相反，试试这个：

def linear_search(intList,target):
    found = False
    count = 0
    #starting while loop
    while count < len(intList):
        if intList[count] == target:
            print("The character is", count, "on the index.")
            found = True
        else:
            count += 1

    if not found:
        print("The character is -1.")

    return count

Answer 2

这是一个使用元组的解决方案：

def linear_search(intList,target):
    found = False
    count = 0
    for i , char in enumerate(sentence):
        if char == target:
            found = True
            print(i , char)
    if not found:
        return -1
    return count

sentence = input('Enter a sentence: ')
character = input('Enter a character: ')
character_found = linear_search(sentence,character)

输出：

Enter a sentence: i have a a time
Enter a character: a
3 a
7 a
9 a
11 a

由于while循环中的break语句，原始代码失败。但是，即使你删除了中断，你也会得到一个无限循环，因为如果找到该字符，你永远不会增加计数器。