查找加起来等于给定字符串的所有子字符串组合

Question

我正在尝试创建一个数据结构来保存所有可能的子字符串组合，这些组合加起来就是原始字符串。例如，如果字符串是 "java"，则有效结果将是 "j", "ava"、"ja", "v", "a"，无效结果将是 "ja", "a" 或 "a", "jav"

我很容易找到所有可能的子字符串

    String string = "java";
    List<String> substrings = new ArrayList<>();
    for( int c = 0 ; c < string.length() ; c++ )
    {
        for( int i = 1 ; i <= string.length() - c ; i++ )
        {
            String sub = string.substring(c, c+i);
            substrings.add(sub);
        }
    }
    System.out.println(substrings);

现在我正在尝试构建一个仅包含有效子字符串的结构。但这并不容易。我在一个非常丑陋的代码的迷雾中，摆弄着索引，并且没有完成的地方，很可能完全走错了路。有什么提示吗？

Answer 1

这是一种方法：

static List<List<String>> substrings(String input) {

    // Base case: There's only one way to split up a single character
    // string, and that is ["x"] where x is the character.
    if (input.length() == 1)
        return Collections.singletonList(Collections.singletonList(input));

    // To hold the result
    List<List<String>> result = new ArrayList<>();

    // Recurse (since you tagged the question with recursion ;)
    for (List<String> subresult : substrings(input.substring(1))) {

        // Case: Don't split
        List<String> l2 = new ArrayList<>(subresult);
        l2.set(0, input.charAt(0) + l2.get(0));
        result.add(l2);

        // Case: Split
        List<String> l = new ArrayList<>(subresult);
        l.add(0, input.substring(0, 1));
        result.add(l);
    }

    return result;
}

输出：

[java]
[j, ava]
[ja, va]
[j, a, va]
[jav, a]
[j, av, a]
[ja, v, a]
[j, a, v, a]

Answer 2

这似乎是找到字符串长度的compositions，并使用这些组合来制作子字符串的问题。所以一个数n有2^n-1个组合，对于长字符串可能会有点耗时...

Answer 3

可能有人想要另一种非递归且不占用内存来保存列表的解决方案：

public static List<List<String>> substrings(final String input) {
    if(input.isEmpty())
        return Collections.emptyList();
    final int size = 1 << (input.length()-1); 
    return new AbstractList<List<String>>() {

        @Override
        public List<String> get(int index) {
            List<String> entry = new ArrayList<>();
            int last = 0;
            while(true) {
                int next = Integer.numberOfTrailingZeros(index >> last)+last+1;
                if(next == last+33)
                    break;
                entry.add(input.substring(last, next));
                last = next;
            }
            entry.add(input.substring(last));
            return entry;
        }

        @Override
        public int size() {
            return size;
        } 
    };
}

public static void main(String[] args) {
    System.out.println(substrings("java"));
}

输出：

[[java], [j, ava], [ja, va], [j, a, va], [jav, a], [j, av, a], [ja, v, a], [j, a, v, a]]

它只是根据其索引计算出下一个组合。

Answer 4

为了防止有人在 python 中寻找相同的算法，这里是 Python 中的实现：

from itertools import combinations

def compositions(s):
    n = len(s)
    for k in range(n):
        for c in combinations(range(1, n), k):
            yield tuple(s[i:j] for i, j in zip((0,) + c, c + (n,)))

工作原理示例：

>>> for x in compositions('abcd'):
...     print(x)
('abcd',)
('a', 'bcd')
('ab', 'cd')
('abc', 'd')
('a', 'b', 'cd')
('a', 'bc', 'd')
('ab', 'c', 'd')
('a', 'b', 'c', 'd')

稍加修改即可生成不同顺序的合成：

def compositions(s):
    n = len(s)
    for k in range(n):
        for c in itertools.combinations(range(n - 1, 0, -1), k):
            yield tuple(s[i:j] for i, j in zip((0,) + c[::-1], c[::-1] + (n,)))

它会给你这个：

>>> for x in compositions('abcd'):
...     print(x)
('abcd',)
('abc', 'd')
('ab', 'cd')
('a', 'bcd')
('ab', 'c', 'd')
('a', 'bc', 'd')
('a', 'b', 'cd')
('a', 'b', 'c', 'd')

还有一个小的补充，你可以只生成指定数量的拆分：

def compositions(s, r=None):
    n = len(s)
    r = range(n) if r is None else [r - 1]
    for k in r:
        for c in itertools.combinations(range(n - 1, 0, -1), k):
            yield tuple(s[i:j] for i, j in zip((0,) + c[::-1], c[::-1] + (n,)))

>>> for x in compositions('abcd', 3):
...     print(x)
('ab', 'c', 'd')
('a', 'bc', 'd')
('a', 'b', 'cd')

Answer 5

只是添加到列表结果的不同递归解决方案

static List<List<String>> substrings(String input) {
    List<List<String>> result = new ArrayList<>();
    if (input.length() == 1) {
        result.add(Arrays.asList(new String[]{input}));
    }
    else {
        //iterate j, ja, jav, jav
        for (int i = 0; i < input.length()-1; i++ ) {
            String root = input.substring(0,i+1);
            String leaf = input.substring(i+1);
            for( List<String> strings: substrings(leaf) ) {
                ArrayList<String> current = new ArrayList<String>();
                current.add(root);
                current.addAll(strings);
                result.add(current);
            }
        }
        //adds the whole string as one of the leaves (ie. java, ava, va, a)
        result.add(Arrays.asList(new String[]{input}));
    }
    return result;
}

查找加起来等于给定字符串的所有子字符串组合

Find all the combination of substrings that add up to the given string

java

string

iteration

algorithm

recursion