2

我正在尝试获取用户输入的所有可能的单词组合。比如单词输入likeaa, bb, cc应该给出

aa
bb
cc
aabb
aacc
bbaa
bbcc
ccaa
ccbb
aabbcc
aaccbb
bbaacc
bbccaa
ccbbaa
ccaabb

它们可以按任何顺序排列，但最好按排序顺序排列。

在过去的一个小时里，我一直在努力做到这一点，我想我对一些我无法弄清楚并继续循环的事情感到困惑。有人可以指出我正确的方向。

到目前为止的代码是

import java.util.Arrays;
import java.util.Scanner;

public class WordCombination {
private static String[] wordlist;
private static String[] wordcomb;

public static void main(String[] argv){
    Scanner a = new Scanner(System.in);

    System.out.print("Enter Words:");
    setWords(a.nextLine());

    creatCombinations();

}

private static void setWords(String words){
    System.out.println("Entered words: " + words);
    wordlist = words.split("\\s+");
    Arrays.sort(wordlist);
    System.out.println(Arrays.toString(wordlist));
}

private static void creatCombinations(){
    int size = wordlist.length;
    int perm = (int) Math.pow(size, size);
    int c = 1;

    wordcomb = new String[perm];

    String word = "";
    /*
    for(int i = 0; i < size; i++){
        word = wordlist[i];         
        for(int j = 0; j < size; j++){
            word += wordlist[j];            
            System.out.println(c + ": " + word + ".com");
            c++;
        }
        word = "";
    }*/
    int l = 0;
    for(int i = 0; i < size; i++){
        int permj = (i+1) * size;
        System.out.println("------------------> " + permj + " permutations for " + (i + 1) + " words combination");
        int iterations = permj * (i+1);
        System.out.println(" Iterations: " + iterations);
        word = wordlist[i];
        for(int j = 0; j < permj; j++){
            for(int k = 0; k < i+1; k++){
                int permi = i * size;
                int index = permi + (k%permj);
                wordcomb[c-1] += "" + wordlist[l%size];
                out(l%size + "");
                l++;
            }
            System.out.println(c + ": " + wordcomb[c-1]);
            c++;
        }
        word = "";
    }

}

private static void out(String s){
    System.out.println(s);
}

}

*编辑：我得到的输出是 *

    Enter Words:1 2 3
Entered words: 1 2 3
[1, 2, 3]
------------------> 3 permutations for 1 words combination
 Iterations: 3
0
1: 1
1
2: 2
2
3: 3
------------------> 6 permutations for 2 words combination
 Iterations: 12

Enter Words:aa bb cc
Entered words: aa bb cc
[aa, bb, cc]
------------------> 3 permutations for 1 words combination
 Iterations: 3
0
1: aa
1
2: bb
2
3: cc
------------------> 6 permutations for 2 words combination
 Iterations: 12
0
1
4: aabb
2
0
5: ccaa
1
2
6: bbcc
0
1
7: aabb
2
0
8: ccaa
1
2
9: bbcc
------------------> 9 permutations for 3 words combination
 Iterations: 27
0
1
2
10: aabbcc
0
1
2
11: aabbcc
0
1
2
12: aabbcc
0
1
2
13: aabbcc
0
1
2
14: aabbcc
0
1
2
15: aabbcc
0
1
2
16: aabbcc
0
1
2
17: aabbcc
0
1
2
18: aabbcc
4

4 回答 4

5

一个问题是它wordcomb[c-1]以 开头null，而不是空字符串。当您添加到它时，null将转换为"null"，其余的将添加到其中。

所以你需要初始化wordcomb[c-1]为一个空字符串：

for(int j = 0; j < permj; j++){
    wordcomb[c-1] = ""; // <--------------------------
    for(int k = 0; k < i+1; k++){
        int permi = i * size;
        int index = permi + (k%permj);
        wordcomb[c-1] += "" + wordlist[l%size];
        out(l%size + "");
        l++;
    }
    System.out.println(c + ": " + wordcomb[c-1]);
    c++;
}
于 2013-08-16T10:29:35.220 回答
3

我认为你的方法从根本上是有缺陷的。在您的输出中，排列的第一个字符 = 前一个字符的最后一个字符 + 1，排列的每个字符 = 前一个字符 + 1，这不是排列的工作方式。所以我认为它不容易被修复。

下面是一种使用递归的工作方式。它只是尝试一个位置的每个字符，然后递归到下一个位置。

import java.util.Arrays;

public class NewMain
{
   public static void main(String[] args)
   {
      c = Arrays.asList("aa","bb","cc").toArray(new String[0]);
      permutation(0);
      System.out.println("Number of permutations = " + count);
   }

   static String[] c;
   static int count = 0;

   static void swap(int pos1, int pos2)
   {
      String temp = c[pos1];
      c[pos1] = c[pos2];
      c[pos2] = temp;
   }

   public static void permutation(int start)
   {
      if (start != 0)
      {
         for (int i = 0; i < start; i++)
            System.out.print(c[i]);
         System.out.println();
         count++;
      }

      for (int i = start; i < c.length; i++)
      {
         swap(start, i);
         permutation(start + 1);
         swap(start, i);
      }
   }
}

印刷：

aa
aabb
aabbcc
aacc
aaccbb
bb
bbaa
bbaacc
bbcc
bbccaa
cc
ccbb
ccbbaa
ccaa
ccaabb
Number of permutations = 15
于 2013-08-16T10:50:35.847 回答
0

这个问题基本上是由打印空字符串对象引起的。尝试检查创建或打印。某处字符串具有空值:)

于 2013-08-16T10:29:58.177 回答
-1

您应该通过调用另一种方法进行排序

Arrays.sort(sortedIDXs, new Comparator<String>() {
 public int compare(String idx1,String idx2) {
    //Your method of sorting here
    //return 1 if string idx1 should be before, 0 if equal and -1 in opposite 
 }
于 2013-08-16T10:34:37.683 回答