algorithm - 生成字符串列表及其子字符串排列的算法

Question

这个算法已经逃离了我一段时间了。假设我得到了字符串“cccaatt”。我正在尝试生成每个重复字母子串的所有可能变体。例如，“cccaatt”作为输入将返回：

猫, 猫, caat, caatt, ccat, ccatt, ccaat, ccaatt, cccat, cccatt, cccaat, cccaatt

结果的顺序无关紧要，只要它返回所有结果即可。一般来说，输入是一个字符串，由g组重复的字母组成，每组k_n个字母长。

我的直觉是，这是一个递归算法，但它的确切结构一直很难理解。

Answer 1

如果您存储每个字母的字母表和最大出现次数（正如评论中所提到的），您可以这样做：

function variations(letter_type, current string) {
    if (letter_type is in the alphabet) {
        while (string has fewer than the max amount of that letter) {
            add one of that letter to current string
            variations(next letter, current string)
        }
    } else {
        print current string // since there are no more letters to add
    }
}

在 Java 中：

public class Variations {

    static String[] alphabet = {"c","a","t"};
    static int[] maximums = {3, 2, 2};

    public static void main(String[] args) {
        variations(0, "");
    }

    public static void variations(int letter_type, String curr) {
        if (letter_type < alphabet.length) {
            for (int i = 1; i <= maximums[letter_type]; i++) {
            curr += alphabet[letter_type];
            variations(letter_type+1, curr);
            }
        } else {
            System.out.println(curr);
        } 
    }

}

Answer 2

将字符串分解为数字列表和重复次数，即“cccaatt” => [(c,3), (a,2), (t,2)]。然后可以递归地定义问题。

Let xs = [(a_1, n_1), (a_2, n_2), (a_3, n_3), ... (a_k, n_k)]
define Perm(xs):
    if len(xs) == 1:
        return all length variations of xs
    else:
        return every sequence in Perm(x[:-1]) appended with one or more from x[-1]

我很快就会有一个 python 示例。

> perm("cccaatt")
> ['cat', 'ccat', 'cccat', 'caat', 'ccaat', 'cccaat', 'catt', 'ccatt', 'cccatt', 'caatt', 'ccaatt', 'cccaatt']

附上代码

def perm(xs):
    if not xs:
        return []

    # group them into the correct format, probably should have used groupby + zip
    l = [(xs[0],1)]
    for x in xs[1:]:
        last, num = l[-1]
        if last == x:
            l[-1] = (last, num+1)
        else:
            l.append((x, 1))

    # print(l)
    print(recurse(l))

# this is where the real work is done.
def recurse(xs):
    if len(xs) == 1:
        return [ xs[0][0] * x for x in range(1, xs[0][1] + 1) ]

    prev = recurse(xs[:-1])
    char, num = xs[-1]
    return [ y + x * char for x in range(1,num + 1) for y in prev ]

Answer 3

Python itertools 模块具有强大的工具来分组，然后迭代导致以下程序的组成员。

我已经展示了一些中间结果并使用 pprint 模块来漂亮地打印答案：

Python 2.7.3 (default, Aug  1 2012, 05:16:07) 
[GCC 4.6.3] on linux2
Type "copyright", "credits" or "license()" for more information.
>>> import itertools
>>> instring = "cccaatt"
>>> [(x[0], list(x[1])) for x in itertools.groupby(instring)]
[('c', ['c', 'c', 'c']), ('a', ['a', 'a']), ('t', ['t', 't'])]
>>> xx = [list(x[0]*n for n in range(1, len(list(x[1]))+1)) for x in itertools.groupby(instring)]
>>> xx
[['c', 'cc', 'ccc'], ['a', 'aa'], ['t', 'tt']]
>>> list(itertools.product(*xx))
[('c', 'a', 't'), ('c', 'a', 'tt'), ('c', 'aa', 't'), ('c', 'aa', 'tt'), ('cc', 'a', 't'), ('cc', 'a', 'tt'), ('cc', 'aa', 't'), ('cc', 'aa', 'tt'), ('ccc', 'a', 't'), ('ccc', 'a', 'tt'), ('ccc', 'aa', 't'), ('ccc', 'aa', 'tt')]
>>> from pprint import pprint as pp
>>> pp(list(itertools.product(*xx)))
[('c', 'a', 't'),
 ('c', 'a', 'tt'),
 ('c', 'aa', 't'),
 ('c', 'aa', 'tt'),
 ('cc', 'a', 't'),
 ('cc', 'a', 'tt'),
 ('cc', 'aa', 't'),
 ('cc', 'aa', 'tt'),
 ('ccc', 'a', 't'),
 ('ccc', 'a', 'tt'),
 ('ccc', 'aa', 't'),
 ('ccc', 'aa', 'tt')]
>>> 

或者作为一个函数：

>>> def stringexpand(instring):
    xx = [list(x[0]*n for n in range(1, len(list(x[1]))+1)) for x in itertools.groupby(instring)]
    return list(itertools.product(*xx))

>>> pp(stringexpand("cccaatt"))
[('c', 'a', 't'),
 ('c', 'a', 'tt'),
 ('c', 'aa', 't'),
 ('c', 'aa', 'tt'),
 ('cc', 'a', 't'),
 ('cc', 'a', 'tt'),
 ('cc', 'aa', 't'),
 ('cc', 'aa', 'tt'),
 ('ccc', 'a', 't'),
 ('ccc', 'a', 'tt'),
 ('ccc', 'aa', 't'),
 ('ccc', 'aa', 'tt')]
>>> 

你似乎需要从它们的部分连接起来的字符串。这可以在这个轻微的 mod 中完成：

def stringexpand(instring):
    xx = [list(x[0]*n for n in range(1, len(list(x[1]))+1)) for x in itertools.groupby(instring)]
    return [''.join(parts) for parts in itertools.product(*xx)]

返回：

['cat',
 'catt',
 'caat',
 'caatt',
 'ccat',
 'ccatt',
 'ccaat',
 'ccaatt',
 'cccat',
 'cccatt',
 'cccaat',
 'cccaatt']