生成字符串排列和组合的智能方法

一尘不染

生成字符串排列和组合的智能方法

java

String database[] = {‘a’, ‘b’, ‘c’};

我想基于给定生成以下字符串序列database。

a
b
c
aa
ab
ac
ba
bb
bc
ca
cb
cc
aaa
...

我只能想到一个漂亮的“虚拟”解决方案。

public class JavaApplication21 {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        char[] database = {'a', 'b', 'c'};

        String query = "a";
        StringBuilder query_sb = new StringBuilder(query);
        for (int a = 0; a < database.length; a++) {
            query_sb.setCharAt(0, database[a]);
            query = query_sb.toString();                    
            System.out.println(query);            
        }

        query = "aa";
        query_sb = new StringBuilder(query);
        for (int a = 0; a < database.length; a++) {
            query_sb.setCharAt(0, database[a]);    
            for (int b = 0; b < database.length; b++) {    
                query_sb.setCharAt(1, database[b]);    
                query = query_sb.toString();                    
                System.out.println(query);
            }
        }

        query = "aaa";
        query_sb = new StringBuilder(query);
        for (int a = 0; a < database.length; a++) {
            query_sb.setCharAt(0, database[a]);    
            for (int b = 0; b < database.length; b++) {    
                query_sb.setCharAt(1, database[b]);    
                for (int c = 0; c < database.length; c++) {                    
                    query_sb.setCharAt(2, database[c]);                        
                    query = query_sb.toString();                    
                    System.out.println(query);
                }
            }
        }
    }
}

解决方案非常愚蠢。从某种意义上说，它是不可扩展的

如果我增加的大小database怎么办？
如果我最终的目标打印字符串长度必须为N怎么办？

是否有任何智能代码可以以一种非常智能的方式生成可缩放的排列和组合字符串？

阅读 193

2020-12-03

共1个答案

一尘不染

您应该检查以下答案：在Java中获取字符串或组合的所有可能排列，包括重复的字符

要获取此代码：

public static String[] getAllLists(String[] elements, int lengthOfList)
{

    //lists of length 1 are just the original elements
    if(lengthOfList == 1) return elements; 
    else {
        //initialize our returned list with the number of elements calculated above
        String[] allLists = new String[(int)Math.pow(elements.length, lengthOfList)];

        //the recursion--get all lists of length 3, length 2, all the way up to 1
        String[] allSublists = getAllLists(elements, lengthOfList - 1);

        //append the sublists to each element
        int arrayIndex = 0;

        for(int i = 0; i < elements.length; i++){
            for(int j = 0; j < allSublists.length; j++){
                //add the newly appended combination to the list
                allLists[arrayIndex] = elements[i] + allSublists[j];
                arrayIndex++;
            }
        }
        return allLists;
    }
}

public static void main(String[] args){
    String[] database = {"a","b","c"};
    for(int i=1; i<=database.length; i++){
        String[] result = getAllLists(database, i);
        for(int j=0; j<result.length; j++){
            System.out.println(result[j]);
        }
    }
}

尽管可以对内存进行进一步的改进，但是由于此解决方案首先生成了所有内存解决方案（数组），然后才能进行打印。但是想法是一样的，那就是使用递归算法。

2020-12-03