所以这里是我写的一个代码,用于查找单词中的回文(检查单词中是否包含单词本身的回文)
条件：字符之间的空格被计算,不被忽略
例如：但是大号是回文,但从技术上讲,由于涉及空间而现在不是.这就是标准.

基于以上所述,以下代码通常应该有效.您可以尝试自己使用不同的测试来检查此代码是否给出任何错误.

def pal(text):
    """

    param text: given string or test
    return: returns index of longest palindrome and a list of detected palindromes stored in temp
    """
    lst = {}
    index = (0, 0)
    length = len(text)
    if length <= 1:
        return index
    word = text.lower()  # Trying to make the whole string lower case
    temp = str()
    for x, y in enumerate(word):
        # Try to enumerate over the word
        t = x
        for i in xrange(x):
            if i != t+1:
                string = word[i:t+1]
                if string == string[::-1]:
                    temp = text[i:t+1]
                    index = (i, t+1)
                    lst[temp] = index
    tat = lst.keys()
    longest = max(tat, key=len)
    #print longest
    return lst[longest], temp

这是一个已经不复存在的版本.我的意思是我试图从中间开始并通过从头开始迭代并通过检查它们是否是相等的字符来检查字符的每个较高和较低的索引来检测回文.如果他们是那么我正在检查它是否像一个常规的回文检查回文.
这就是我所做的

def pal(t):
    text = t.lower()
    lst = {}
    ptr = ''
    index = (0, 0)
    #mid = len(text)/2
    #print mid
    dec = 0
    inc = 0
    for mid, c in enumerate(text):
        dec = mid - 1
        inc = mid + 1
        while dec != 0 and inc != text.index(text[-1]):
            print 'dec {}, inc {},'.format(dec, inc)
            print 'text[dec:inc+1] {}'.format(text[dec:inc+1])
            if dec<0:
                dec = 0
            if inc > text.index(text[-1]):
                inc = text.index(text[-1])
            while text[dec] != text[inc]:
                flo = findlet(text[inc], text[:dec])
                fhi = findlet(text[dec], text[inc:])
                if len(flo) != 0 and len(fhi) != 0 and text[flo[-1]] == text[fhi[0]]:
                    dec = flo[-1]
                    inc = fhi[0]
                    print ' break if'
                    break
                elif len(flo) != 0 and text[flo[-1]] == text[inc]:
                    dec = flo[-1]
                    print ' break 1st elif'
                    break
                elif len(fhi) != 0 and text[fhi[0]] == text[inc]:
                    inc = fhi[0]
                    print ' break 2nd elif'
                    break
                else:
                    dec -= 1
                    inc += 1
                    print ' break else'
                    break
            s = text[dec:inc+1]
            print ' s {} '.format(s)
            if s == s[::-1]:
                index = (dec, inc+1)
                lst[s] = index
            if dec > 0:
                dec -= 1
            if inc < text.index(text[-1]):
                inc += 1
    if len(lst) != 0:
        val = lst.keys()
        longest = max(val, key = len)
        return lst[longest], longest, val
    else:
        return index

findlet()有趣：

def findlet(alpha, string):
    f = [i for i,j in enumerate(string) if j == alpha]
    return f

有时它有效：

pal('madem')
dec -1, inc 1,
text[dec:inc+1] 
 s m 
dec 1, inc 3,
text[dec:inc+1] ade
 break 1st elif
 s m 
dec 2, inc 4,
text[dec:inc+1] dem
 break 1st elif
 s m 
dec 3, inc 5,
text[dec:inc+1] em
 break 1st elif
 s m 
Out[6]: ((0, 1), 'm', ['m'])

pal('Avid diva.')
dec -1, inc 1,
text[dec:inc+1] 
 break 2nd if
 s avid div 
dec 1, inc 3,
text[dec:inc+1] vid
 break else
 s avid  
dec 2, inc 4,
text[dec:inc+1] id 
 break else
 s vid d 
dec 3, inc 5,
text[dec:inc+1] d d
 s d d 
dec 2, inc 6,
text[dec:inc+1] id di
 s id di 
dec 1, inc 7,
text[dec:inc+1] vid div
 s vid div 
dec 4, inc 6,
text[dec:inc+1]  di
 break 1st elif
 s id di 
dec 1, inc 7,
text[dec:inc+1] vid div
 s vid div 
dec 5, inc 7,
text[dec:inc+1] div
 break 1st elif
 s vid div 
dec 6, inc 8,
text[dec:inc+1] iva
 break 1st elif
 s avid diva 
dec 8, inc 10,
text[dec:inc+1] a.
 break else
 s va. 
dec 6, inc 10,
text[dec:inc+1] iva.
 break else
 s diva. 
dec 4, inc 10,
text[dec:inc+1]  diva.
 break else
 s d diva. 
dec 2, inc 10,
text[dec:inc+1] id diva.
 break else
 s vid diva. 
Out[9]: ((0, 9), 'avid diva', ['avid diva', 'd d', 'id di', 'vid div'])

并根据我已经提出的标准/条件：

pal('A car, a man, a maraca.')
dec -1, inc 1,
text[dec:inc+1] 
 break else
 s  
dec -1, inc 3,
text[dec:inc+1] 
 s a ca 
dec 1, inc 3,
text[dec:inc+1]  ca
 break if
 s a ca 
dec 2, inc 4,
text[dec:inc+1] car
 break else
 s  car, 
dec 3, inc 5,
text[dec:inc+1] ar,
 break else
 s car,  
dec 1, inc 7,
text[dec:inc+1]  car, a
 break 1st elif
 s a car, a 
dec 4, inc 6,
text[dec:inc+1] r, 
 break 1st elif
 s  car,  
dec 5, inc 7,
text[dec:inc+1] , a
 break 1st elif
 s ar, a 
dec 2, inc 8,
text[dec:inc+1] car, a 
 break 1st elif
 s  car, a  
dec 6, inc 8,
text[dec:inc+1]  a 
 s  a  
dec 5, inc 9,
text[dec:inc+1] , a m
 break else
 s r, a ma 
dec 3, inc 11,
text[dec:inc+1] ar, a man
 break else
 s car, a man, 
dec 1, inc 13,
text[dec:inc+1]  car, a man, 
 s  car, a man,  
dec 7, inc 9,
text[dec:inc+1] a m
 break else
 s  a ma 
dec 5, inc 11,
text[dec:inc+1] , a man
 break else
 s r, a man, 
dec 3, inc 13,
text[dec:inc+1] ar, a man, 
 break if
 s   
dec 8, inc 10,
text[dec:inc+1]  ma
 break if
 s  
dec 6, inc 4,
text[dec:inc+1] 
 break 1st elif
 s r 
dec 3, inc 5,
text[dec:inc+1] ar,
 break else
 s car,  
dec 1, inc 7,
text[dec:inc+1]  car, a
 break 1st elif
 s a car, a 
dec 9, inc 11,
text[dec:inc+1] man
 break else
 s  man, 
dec 7, inc 13,
text[dec:inc+1] a man, 
 break if
 s  
dec 5, inc 2,
text[dec:inc+1] 
 break 1st elif
 s c 
dec 1, inc 3,
text[dec:inc+1]  ca
 break if
 s a ca 
dec 10, inc 12,
text[dec:inc+1] an,
 break 1st elif
 s , a man, 
dec 4, inc 13,
text[dec:inc+1] r, a man, 
 break 1st elif
 s  car, a man,  
dec 11, inc 13,
text[dec:inc+1] n, 
 break 1st elif
 s  man,  
dec 7, inc 14,
text[dec:inc+1] a man, a
 s a man, a 
dec 6, inc 15,
text[dec:inc+1]  a man, a 
 s  a man, a  
dec 5, inc 16,
text[dec:inc+1] , a man, a m
 break else
 s r, a man, a ma 
dec 3, inc 18,
text[dec:inc+1] ar, a man, a mar
 break else
 s car, a man, a mara 
dec 1, inc 20,
text[dec:inc+1]  car, a man, a marac
 break else
 s a car, a man, a maraca 
dec 12, inc 14,
text[dec:inc+1] , a
 break 1st elif
 s an, a 
dec 9, inc 15,
text[dec:inc+1] man, a 
 break if
 s  
dec 7, inc 2,
text[dec:inc+1] 
 break 1st elif
 s c 
dec 1, inc 3,
text[dec:inc+1]  ca
 break if
 s a ca 
dec 13, inc 15,
text[dec:inc+1]  a 
 s  a  
dec 12, inc 16,
text[dec:inc+1] , a m
 break 1st elif
 s man, a m 
dec 8, inc 17,
text[dec:inc+1]  man, a ma
 break 1st elif
 s a man, a ma 
dec 6, inc 18,
text[dec:inc+1]  a man, a mar
 break 1st elif
 s r, a man, a mar 
dec 3, inc 19,
text[dec:inc+1] ar, a man, a mara
 s ar, a man, a mara 
dec 2, inc 20,
text[dec:inc+1] car, a man, a marac
 s car, a man, a marac 
dec 1, inc 21,
text[dec:inc+1]  car, a man, a maraca
 break 1st elif
 s a car, a man, a maraca 
dec 14, inc 16,
text[dec:inc+1] a m
 break 1st elif
 s man, a m 
dec 8, inc 17,
text[dec:inc+1]  man, a ma
 break 1st elif
 s a man, a ma 
dec 6, inc 18,
text[dec:inc+1]  a man, a mar
 break 1st elif
 s r, a man, a mar 
dec 3, inc 19,
text[dec:inc+1] ar, a man, a mara
 s ar, a man, a mara 
dec 2, inc 20,
text[dec:inc+1] car, a man, a marac
 s car, a man, a marac 
dec 1, inc 21,
text[dec:inc+1]  car, a man, a maraca
 break 1st elif
 s a car, a man, a maraca 
dec 15, inc 17,
text[dec:inc+1]  ma
 break 1st elif
 s a ma 
dec 13, inc 18,
text[dec:inc+1]  a mar
 break 1st elif
 s r, a man, a mar 
dec 3, inc 19,
text[dec:inc+1] ar, a man, a mara
 s ar, a man, a mara 
dec 2, inc 20,
text[dec:inc+1] car, a man, a marac
 s car, a man, a marac 
dec 1, inc 21,
text[dec:inc+1]  car, a man, a maraca
 break 1st elif
 s a car, a man, a maraca 
dec 16, inc 18,
text[dec:inc+1] mar
 break 1st elif
 s r, a man, a mar 
dec 3, inc 19,
text[dec:inc+1] ar, a man, a mara
 s ar, a man, a mara 
dec 2, inc 20,
text[dec:inc+1] car, a man, a marac
 s car, a man, a marac 
dec 1, inc 21,
text[dec:inc+1]  car, a man, a maraca
 break 1st elif
 s a car, a man, a maraca 
dec 17, inc 19,
text[dec:inc+1] ara
 s ara 
dec 16, inc 20,
text[dec:inc+1] marac
 break 1st elif
 s car, a man, a marac 
dec 1, inc 21,
text[dec:inc+1]  car, a man, a maraca
 break 1st elif
 s a car, a man, a maraca 
dec 18, inc 20,
text[dec:inc+1] rac
 break 1st elif
 s car, a man, a marac 
dec 1, inc 21,
text[dec:inc+1]  car, a man, a maraca
 break 1st elif
 s a car, a man, a maraca 
dec 19, inc 21,
text[dec:inc+1] aca
 s aca 
dec 21, inc 23,
text[dec:inc+1] a.
 break else
 s ca. 
dec 19, inc 23,
text[dec:inc+1] aca.
 break else
 s raca. 
dec 17, inc 23,
text[dec:inc+1] araca.
 break else
 s maraca. 
dec 15, inc 23,
text[dec:inc+1]  maraca.
 break else
 s a maraca. 
dec 13, inc 23,
text[dec:inc+1]  a maraca.
 break else
 s , a maraca. 
dec 11, inc 23,
text[dec:inc+1] n, a maraca.
 break else
 s an, a maraca. 
dec 9, inc 23,
text[dec:inc+1] man, a maraca.
 break else
 s  man, a maraca. 
dec 7, inc 23,
text[dec:inc+1] a man, a maraca.
 break else
 s  a man, a maraca. 
dec 5, inc 23,
text[dec:inc+1] , a man, a maraca.
 break else
 s r, a man, a maraca. 
dec 3, inc 23,
text[dec:inc+1] ar, a man, a maraca.
 break else
 s car, a man, a maraca. 
dec 1, inc 23,
text[dec:inc+1]  car, a man, a maraca.
 break else
 s a car, a man, a maraca. 
Out[8]: ((13, 16), ' a ', ['', ' a ', 'c', ' ', 'aca', 'ara', 'r'])

有时,它根本不起作用：

    pal('madam')
    dec -1, inc 1,
    text[dec:inc+1] 
     s m 
    dec 1, inc 3,
    text[dec:inc+1] ada
     break 1st elif
     s m 
    dec 2, inc 4,
    text[dec:inc+1] dam
     break 1st elif
     s m 
    dec 3, inc 5,
    text[dec:inc+1] am
     break 1st elif
     s m 
    Out[5]: ((0, 1), 'm', ['m'])

现在考虑女士是一个非常好的回文它应该工作,有很多情况,我没有测试自己,以找出它没有检测到的其他合法的回文.

Q1：为什么有时候没有检测到？

Q2：我想优化我的第二个代码.有什么投入？

问题3：对于比我的第一个代码多得多次迭代的代码,有什么更好的方法呢？

解决方法:

你的解决方案对我来说似乎有点复杂.只需查看所有可能的子字符串并单独检查它们：

def palindromes(text):
    text = text.lower()
    results = []

    for i in range(len(text)):
        for j in range(0, i):
            chunk = text[j:i + 1]

            if chunk == chunk[::-1]:
                results.append(chunk)

    return text.index(max(results, key=len)), results

text.index()只会找到最长的回文结构,所以如果你想要最后一个,请用text.rindex()替换它.