Question

我正在大学里学习Python，我们的任务是检查当前字符串是否几乎是回文。

几乎是回文的字符串是一个字符串，如果您从其中删除一个字符，无论是哪个字符，也只有一个字符，您都可以从左至右或从右至左以相同的方式读取该字符串；例如：abba，aa和aba。

提交代码时，会有一个自动系统检查您的算法，并且系统会告诉我算法存在一些问题，而我找不到问题所在。

这是我函数的代码：

def question3(str):#define function with input string
    i=0#define counter which will run through the whole string from the begining
    j=0#define counter which will run through the whole string without the char we deleted from the begining of the string
    k=0#define counter which will run through the string without the deleted char from the end of the string
    answer=False #define answer

    while i<len(str):#run a loop through the string
        k=len(str)-1
        while j<len(str):# run a loop through the string without the deleted char
            if i==j:# if j is in the place of the deleted chart,j skip to the next char
                j=j+1
                continue
            if k==i:# if k is in the place of the deleted chart,k skip to the next char
                k=k-1
                continue

            if  str[j]==str[k]:#check if the chart in the j's place equal to the char in the k's place
                answer=True

            else:
                answer=False#if the answer is false,we don't need to check the rest of the string because it's already not almost a polyndrom
                break#exit the inner loop

            j=j+1#j moves to the next chart
            k=k-1# k moves to the next chart
        if answer==True:# if we found that the string is almost a polyndrom without a specific char,
            #we don't need to check the rest of the string without the rest of the chars and we can exit the outer loop
            break# exit the loop
        i=i+1# move to the next chart
        j=0#nullify the counter that runs through the whole string from the beginning without a specific char

    print(answer)
    return;

Answer 1

您的代码对于手头的练习似乎有点复杂。这是一个更简单的版本（我认为）。

要检查python中的回文，最简单的方法是这样（来自this answer）：

def check_for_palindrome(s):
    return s == s[::-1]

显然，这没有短路功能，因此对于非常长的字符串，可以将其实现为更快，但是速度很可能不是您分配的要求。

现在，我不能完全理解您的问题，因为可以通过两种不同的方式来理解它。从原始字符串中删除一个字符后：

您正在检查所有可能的新字符串中的至少一个是回文：

def is_almost_palindrome_v1a(s):
    """
    'True' if there is at least one option that is palindrome.
    """
    for i in range(len(s)):
        new_s = s[:i] + s[i+1:]
        is_palindrome = check_for_palindrome(new_s)
        print(new_s, is_palindrome)

        if is_palindrome:
            return True

    return False

这可以使用any()来缩短：

def is_almost_palindrome_v1b(s):
    return any(
        check_for_palindrome(s[:i] + s[i+1:])
        for i in range(len(s)))

您正在检查所有可能的新字符串都是回文

def is_almost_palindrome_v2a(s):
    """
    'False' if there is at least one option that is NOT a palindrome.
    """
    for i in range(len(s)):
        new_s = s[:i] + s[i+1:]
        is_palindrome = check_for_palindrome(new_s)
        print(new_s, is_palindrome)

        if not is_palindrome:
            return False

    return True

这可以使用all()来缩短：

def is_almost_palindrome_v2b(s):
    return all(
        check_for_palindrome(s[:i] + s[i+1:])
        for i in range(len(s)))

Answer 2

谢谢您的评论。有人提到我是第一种方式。我们根本不允许使用任何内置库。我将代码更改为str == str [::-1]，以检查字符串是否是回文。这是我的新代码：

def question3(str):#define function with input string
 strCopy = ""  # define an empty string to copy a string without a char we want to 
delete
    i = 0  # define counter which will run through the original string
    j = 0  # define first help counter
    answer = False  # define answer
    while i < len(str):  # run a loop through the original string
        while j < len(
                str):  # a loop where the original string will be copied to a new string without the deleted char
            if i == j:  # check if the current char is the char we want to delete,if yes,skip to the next char
                j = j + 1
                continue
            strCopy = strCopy + str[j]  # add the char in the j'th place from the original string to the new string
            j = j + 1#moving to the next char
        if strCopy==strCopy[::-1]:#check if the string without the deleted char is a palindrome,if yes,so the original string is almost a palindrome
            answer=True
            break

        i = i + 1#moving to the next char
        strCopy = ""#nullify the string without the deleted char
        j = 0#nullify j

    print(answer)  # print the result(True/False)
    return;

测试几乎回文

2 个答案: