具有回溯的数独算法 - java

Question

所以我有这个大学任务解决数独...我读了算法X和跳舞算法,但他们没有帮助我.

我需要回溯.我用维基百科给出的位置上的数字硬编码了二维数组中的一些索引(所以我确信它是可解的).

我得到的代码如下:

public void solveSudoku(int row, int col)
   {
      // clears the temporary storage array that is use to check if there are
      // dublicates on the row/col
      for (int k = 0; k < 9; k++)
      {
         dublicates[k] = 0;
      }
      // checks if the index is free and changes the input number by looping
      // until suitable
      if (available(row, col))
      {
         for (int i = 1; i < 10; i++)
         {
            if (checkIfDublicates(i) == true)
            {
               board[row][col] = i;
               if (row == 8)
                  solveSudoku(0, col + 1);
               else if (col == 8)
                  solveSudoku(row + 1, 0);
               else
                  solveSudoku(row, col + 1);

               board[row][col] = 0;
            }
         }
      }
      // goes to the next row/col
      else
      {
         if (row == 8)
            solveSudoku(0, col + 1);
         else if (col == 8)
            solveSudoku(row + 1, 0);
         else
            solveSudoku(row, col + 1);
      }
   }

   /**
    * Checks if the spot on the certain row-col index is free of element
    * 
    * @param row
    * @param col
    * @return
    */
   private boolean available(int row, int col)
   {
      if (board[row][col] != 0)
         return false;
      else
         return true;
   }

   /**
    * Checks if the number given is not already used in this row/col
    * 
    * @param numberToCheck
    * @return
    */
   private boolean checkIfDublicates(int numberToCheck)
   {
      boolean temp = true;
      for (int i = 0; i < dublicates.length; i++)
      {
         if (numberToCheck == dublicates[i])
         {
            temp = false;
            return false;
         }
         else if (dublicates[i] == 0)
         {
            dublicates[i] = numberToCheck;
            temp = true;
            return true;
         }
      }
      return temp;
   }

我正在使用StackOverflow

// goes to the next row/col
          else
          {
             if (row == 8)
                solveSudoku(0, col + 1);
             else if (col == 8)
                solveSudoku(row + 1, 0);
             else
                solveSudoku(row, col + 1);
          }

这意味着我必须在某个时候停止递归,但我无法弄清楚如何!如果您在solve()功能中发现任何其他错误- 请告诉我.因为我不确定我完全理解"回溯"的事情......

Answer 1

例如，如果您跟踪当前的递归深度，您可以停止递归

public void solveSudoku(int row, int col, int recursionDepth) {
    // get out of here if too much
    if (recursionDepth > 15) return;

    // regular code...
    // at some point call self with increased depth
    solveSudoku(0, col + 1, recursionDepth + 1);
}

如果您在solve()函数中发现任何其他错误，请告诉我。

代码太多了:)