JAVA中字符数组中的特定元素排列？

Question

如何列出字符数组中指定的任何字母的所有大小写排列？ 假设我有一个这样的字符数组：['h'，'e'，'l'，'l'，'o'] 我想打印出字母“l”的可能组合，这样它就能打印出来[你好，你好，你好，你好]

这就是我到目前为止所拥有的（唯一的问题是我可以打印排列，但是我无法在实际单词中打印它们。因此我的代码打印[ll，ll，ll，ll]，而不是上面的示例

我的代码：

import java.util.ArrayList;
import java.util.HashSet;

public class Main {

public static void main(String[] args) {
    //Sample Word
    String word = "Tomorrow-Today";

    //Sample Letters for permutation
    String rule_char_set = "tw";


    ArrayList<Character> test1 = lettersFound(word, rule_char_set);

    printPermutations(test1);







}

public static void printPermutations(ArrayList<Character> arrayList) {
    char[] chars = new char[arrayList.size()];
    int charIterator = 0;

    for(int i=0; i<arrayList.size(); i++){
        chars[i] = arrayList.get(i);
    }

    for (int i = 0, n = (int) Math.pow(2, chars.length); i < n; i++) {
        char[] permutation = new char[chars.length];
        for (int j =0; j < chars.length; j++) {
            permutation[j] = (isBitSet(i, j)) ? Character.toUpperCase(chars[j]) : chars[j];
        }
        System.out.println(permutation);
    }
}

public static boolean isBitSet(int n, int offset) {
    return (n >> offset & 1) != 0;
}

public static ArrayList<Character> lettersFound(String word, String rule_char_set) {

    //Convert the two parameter strings to two character arrays
    char[] wordArray = word.toLowerCase().toCharArray();
    char[] rule_char_setArray = rule_char_set.toLowerCase().toCharArray();

    //ArrayList to hold found characters;
    ArrayList<Character> found = new ArrayList<Character>();

    //Increments the found ArrayList that stores the existent values.
    int foundCounter = 0;


    for (int i = 0; i < rule_char_setArray.length; i++) {
        for (int k = 0; k < wordArray.length; k++) {
            if (rule_char_setArray[i] == wordArray[k]) {
                found.add(foundCounter, rule_char_setArray[i]);
                foundCounter++;

            }
        }
    }
    //Convert to a HashSet to get rid of duplicates
    HashSet<Character> uniqueSet = new HashSet<>(found);

    //Convert back to an ArrayList(to be returned) after filtration of duplicates.
    ArrayList<Character> filtered = new ArrayList<>(uniqueSet);

    return filtered;
}

}

Answer 1

您需要在程序中做一些更改。您的逻辑是完美的，您需要首先找到给定单词中要更改的characters。找到它们后，查找characters中的powerset以打印所有排列，但这将只打印给定单词中存在的rule-char-set字符中的permuatation

您需要做的一些更改是，首先查找word中包含rule-char-set字符的所有indexes。然后找到存储在ArrayList中的所有subsets索引，然后对于每个子集的每个元素，使character出现在index的大写字母上，这将提供您所需的所有permutation

考虑一个例子，{{CD14>}和^ {CD15>}，然后首先需要在字符串{{CD7}}中找到所有索引^ {CD16>}和^ {CD17>}。

所以这里的索引是0,2,3。将其存储在ArrayList中，然后查找其powerset。然后，对于每个subset，使character出现在index上的uppercase字母

Word[] = {'h','e','l','l','o'}
indexes =  0 , 1 , 2 , 3 , 4


index[]= { 0 , 2 ,3}  //Store the indexes of characters which are to be changed


BITSET      |       SUBSET      |       word

000         |         -         |       hello 
001         |        {3}        |       helLo 
010         |        {2}        |       heLlo 
011         |       {2,3}       |       heLLo 
100         |        {0}        |       Hello 
101         |       {0,3}       |       HelLo 
110         |       {0,2}       |       HeLlo 
111         |      {0,2,3}      |       HeLLo 

代码：

import java.util.ArrayList;
import java.util.HashSet;

public class Main {

    public static void main(String[] args) {
        //Sample Word
        String word = "Tomorrow-Today";

        //Sample Letters for permutation
        String rule_char_set = "tw";


        ArrayList<Integer> test1 = lettersFound(word, rule_char_set); //To store the indexes of the characters

        printPermutations(word,test1);

    }

    public static void printPermutations(String word,ArrayList<Integer> arrayList) {
        char word_array[]=word.toLowerCase().toCharArray();
        int length=word_array.length;
        int index[]=new int[arrayList.size()];

        for(int i=0; i<arrayList.size(); i++){
            index[i] = arrayList.get(i);
        }

        for (int i = 0, n = (int) Math.pow(2, index.length); i < n; i++) {
            char[] permutation = new char[length];

            System.arraycopy(word_array,0,permutation,0,length);    

            //First copy the original array and change 
            //only those character whose indexes are present in subset

            for (int j =0; j < index.length; j++) {
                permutation[index[j]] = (isBitSet(i, j)) ? Character.toUpperCase(permutation[index[j]]) : permutation[index[j]];
            }
            System.out.println(permutation);
        }
    }

    public static boolean isBitSet(int n, int offset) {
        return (n >> offset & 1) != 0;
    }

    public static ArrayList<Integer> lettersFound(String word, String rule_char_set) {

        //Convert the two parameter strings to two character arrays
        char[] wordArray = word.toLowerCase().toCharArray();
        char[] rule_char_setArray = rule_char_set.toLowerCase().toCharArray();

        //ArrayList to hold found characters;
        ArrayList<Integer> found = new ArrayList<Integer>();

        //Increments the found ArrayList that stores the existent values.
        int foundCounter = 0;


        for (int i = 0; i < rule_char_setArray.length; i++) {
            for (int k = 0; k < wordArray.length; k++) {
                if (rule_char_setArray[i] == wordArray[k]) {
                    found.add(foundCounter, k);       //Store the index of the character that matches
                    foundCounter++;

                }
            }
        }
        return found;
    }

}

输出：

tomorrow-today
Tomorrow-today
tomorrow-Today
Tomorrow-Today
tomorroW-today
TomorroW-today
tomorroW-Today
TomorroW-Today

Answer 2

Sanket Makani的答案是完美的

我可以提供一个更客观的方法来解决这个问题

作为输入，您需要修改字符串和字符，这些字符应替换为已修改的大小写（大写或小写）。 作为输出，您将拥有所有置换字符串

我将创建一个包含索引的结构，以及可能随以下内容而更改的值：

class Change {
int index;
char values[];
}

我们将需要进行所有可能的组合，因此让我们包括一个字段，该字段将告诉我们的结构中当前使用的字符，并添加一些方法：

class Change {
int index;
char values[];
int cur;

void reset() {cur=0;}
boolen isMax(){return cur==values.length-1;}
void next(){cur++;}
char getValue(){ return values[cur]; }
} 

我们将有一个这些类的列表或数组，我们将把它们放入一个单独的类中

class Combination {
Change changes[];

  void reset() { for (Change c: changes) c.reset();}

  boolean next() {
   for ( int i=0; i<changes.length; i++)
      if ( changes[i].isMax())
         changes[i].reset(); // next change will be taken in cycle, with "next()"
      else {changes[i].next(); return true;}

   return false; // all changes are max
  }
}

所以，当您通过输入数据初始化“组合”类时，您可以在循环中使用它

Combination c = new Combination();
.... // initialization here 
c.reset();
do {
 ... //  update and print your string
} while ( c.next()  );

初始化“组合”并使用值更新我在您之后留下的输入字符串：）

Answer 3

对于置换的情况，我认为递归在可读性方面是最合适的，考虑到它在性能方面可能不是最好的

我的做法是：

public static void main(String[] args) {
    generateCombinations("hello", "l", "");
}

public static void generateCombinations(String text, String changingLetters, String current) {
    if (0 == text.length()) {
        System.out.println(current);
        return;
    }
    String currentLetter = text.substring(0, 1);
    if (changingLetters.contains(currentLetter)) {
        generateCombinations(text.substring(1), changingLetters, current + currentLetter.toUpperCase());
    }
    generateCombinations(text.substring(1), changingLetters, current + currentLetter);
}

主执行的输出将是：

heLLo
heLlo
helLo
hello