查找加起来为给定字符串的所有子字符串组合

本文介绍了查找加起来为给定字符串的所有子字符串组合的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我正在尝试创建一个数据结构，它包含所有可能的子字符串组合，这些组合加起来是原始字符串.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"

I'm trying to create a data structure that holds all the possible substring combinations that add up to the original string. For example, if the string is "java" the valid results would be "j", "ava", "ja", "v", "a", an invalid result would be "ja", "a" or "a", "jav"

我很容易找到所有可能的子字符串

I had it very easy in finding all the possible substrings

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.但它几乎没有那么容易.我陷入了一个非常丑陋的代码的迷雾中，摆弄着索引，而且还没有完成，很可能完全走错了路.有什么提示吗?

and now I'm trying to construct a structure that holds only the valid substrings. But its not nearly as easy. I'm in the mist of a very ugly code, fiddling around with the indexes, and no where near of finishing, most likely on a wrong path completely. Any hints?

推荐答案

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]