Python:搜索一个单词中最长的回文和一个单词/字符串中的回文

Question

所以这是我编写的代码，用于查找单词中的回文(检查单词中是否有回文，包括单词本身)条件:字符之间的空格被计算在内，不会被忽略示例:A but tuba 是回文，但从技术上讲，由于涉及的空间现在不是了。这就是标准。

基于以上，下面的代码通常应该可以工作。您可以自己尝试不同的测试来检查这段代码是否有任何错误。

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

这是它的一个失效版本。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字符是否相等来检查字符的每个较高和较低索引来检测回文。如果是，那么我正在检查它是否像常规回文检查一样是回文。这是我所做的

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() 的乐趣:

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'])

根据我提出的标准/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'])

现在考虑到 madam 是一个非常好的回文，它应该可以工作，并且有很多情况我还没有测试自己以找出它没有检测到的其他合法回文。

Q1:为什么有时检测不到？

Q2:我想为此优化我的第二个代码。有什么意见吗？

问题 3:有什么比我的第一个迭代多次的代码更有效的代码的更好方法？

Answer 1

您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() 只会找到最长回文的第一次出现，所以如果您想要最后一次出现，请将其替换为 text.rindex()。