填充二维数组时如何生成唯一字符？

Question

如何做到这一点，以便在运行代码时输出网格时，没有两个数字或字母会相同？当我当前运行此代码时，我可以获得 3x“L”或 2x“6”，我该如何让它只出现一次？

package polycipher;

import java.util.ArrayList;

public class Matrix {
private char[][] matrix = new char[6][6];
private int[] usedNumbers = new int[50];

{for(int x = 0; x < usedNumbers.length; x++) usedNumbers[x] = -1;}

private final char[] CIPHER_KEY = {'A','D','F','G','V','X'};
private final String validChars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";

public Matrix() {
    int random;
    for(int i = 0; i < CIPHER_KEY.length; i++) {
        for(int j = 0; j < CIPHER_KEY.length; j++) {
            validation: while(true) {
                random = (int)(Math.random()*validChars.length()-1);
                for(int k = 0; k < usedNumbers.length; k++) {
                    if(random == usedNumbers[k]) continue validation;
                    else if(usedNumbers[k]==-1) usedNumbers[k] = random;
                }
                break;
            }
            matrix[i][j] = validChars.split("")[random].charAt(0);  
        }
    }
}

public String toString() {
    String output = "   A D F G V X\n";
    for(int i = 0; i < CIPHER_KEY.length; i++) {
        output += CIPHER_KEY[i] + "  ";
        for(int j = 0; j < CIPHER_KEY.length; j++) {
            output += matrix[i][j] + " ";
        }
        output += "\n";
    }
    return output;
}
}

Answer 1

这应该比验证每个随机选择要快得多：

1. 将您的有效字符存储到一个数组中；

   char[] valid = validChars.toCharArray();
   

2. Shuffle the array;

   shuffle(valid)
   

3. 遍历矩阵中的位置，按照它们在打乱数组中出现的相同顺序存储元素。

   assert (CIPHER_KEY.length * CIPHER_KEY.length) <= valid.length;
   int k = 0;
   for (int i = 0; i < CIPHER_KEY.length; i++) {
       for (int j = 0; j < CIPHER_KEY.length; j++) {
           matrix[i][j] = valid[k++];
       }
   }
   

Answer 2

如果旧随机数在地图中，则使用一个集合并生成一个新随机数： 伪代码：

Set<Integer> set = new HashSet<Integer>();
for () {
    int random = (int)(Math.random()*validChars.length()-1);
    //Your code for validation here (move it to a function)
    while (!set.contains(random)){
         int random = (int)(Math.random()*validChars.length()-1);
         //Your code for validation here (move it to a function)
    }
    //If we exit this loop it means the set doesn't contain the number
    set.add(random);
    //Insert your code here
}