leetcode-37 Sudoku Solver

解决数独谜题算法

最新推荐文章于 2021-10-02 00:35:03 发布

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最新推荐文章于 2021-10-02 00:35:03 发布

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本文介绍了一种通过编程解决数独谜题的方法，确保每行、每列及每个3x3子格内的数字1-9各出现一次。采用暴力搜索策略，递归地尝试填充空格，直至找到唯一解。

Write a program to solve a Sudoku puzzle by filling the empty cells.

A sudoku solution must satisfy all of the following rules:

Each of the digits 1-9 must occur exactly once in each row.
Each of the digits 1-9 must occur exactly once in each column.
Each of the the digits 1-9 must occur exactly once in each of the 9 3x3 sub-boxes of the grid.

Empty cells are indicated by the character '.'.

A sudoku puzzle...

...and its solution numbers marked in red.

Note:

The given board contain only digits 1-9 and the character '.'.
You may assume that the given Sudoku puzzle will have a single unique solution.
The given board size is always 9x9.

暴力解法对每个.的单元格进行可填元素遍历直到符合条件

   Map <Character,Character> all = new HashMap<>();
   {
       all.put('1','1');
       all.put('2','2');
       all.put('3','3');
       all.put('4','4');
       all.put('5','5');
       all.put('6','6');
       all.put('7','7');
       all.put('8','8');
       all.put('9','9');
   }
   //在空的*中填写数字并让他符合要求的条件
public void solveSudoku(char[][] board) {
   isSolveSudoku(board,0,0);
}

public boolean isSolveSudoku(char[][] board,int beginIndex,int jIndex) {
for(int i =0;i<board[0].length;i++) {
   for(int j=0;j<board.length;j++) {
       if(board[i][j] == '.') {
           if(board[6][0] == '9') {
               System.out.println();
           }
           if(i==8 && j==6) {
               System.out.println();
           }
           List<Character> tmp = getValid(board,i,j);
           if(tmp.isEmpty()) {
               return false;
           }
           for(Character ad:tmp) {
               board[i][j] = ad;
               if(!isSolveSudoku(board,i,j)) {
                   board[i][j] = '.';
               }else {
                   return true;
               }
           }
           if(board[i][j] == '.') {
               return false;
           }else {
               return true;
           }

       }
   }
}

return true;
}

public List<Character> getValid(char[][] board,int i,int j){
       Map<Character, Character> tmpMap = new HashMap<>(all);

       for (int tmp = 0; tmp < board.length; tmp++) {
           if (Character.isDigit(board[tmp][j])) {
               tmpMap.remove(board[tmp][j]);
           }
       }

       for (int tmp = 0; tmp < board[i].length; tmp++) {
           if (Character.isDigit(board[i][tmp])) {
               tmpMap.remove(board[i][tmp]);
           }
       }

       int left = (i / 3)*3;
       int right = (j / 3)*3;
       for (int k = left; k < left + 3; k++) {
           for (int l = right; l < right + 3; l++) {
               if (Character.isDigit(board[k][l])) {
                   tmpMap.remove(board[k][l]);
               }

           }
       }
       return new ArrayList<>(tmpMap.keySet());

}