原题

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

A sudoku solution must satisfy all of the following rules:

1. Each of the digits 1-9 must occur exactly once in each row.
2. Each of the digits 1-9 must occur exactly once in each column.
3. 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 '.'.

求解 解答

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.

解法

数独游戏是一个经典的游戏，每一行、每一列和每一个小九格中填入1~9不重复的数字。

1. 使用序号i=0~80遍历九宫格，再用y = i / 9; x = i % 9;转换成坐标。一般使用x和y坐标也可以，不过这样需要两重循环，递归起来也麻烦一点。
2. 判断同一行、同一列中的数字都简单，判断同在一个小九格中有一个技巧，先计算xx = x / 3 * 3, yy = y / 3 * 3;，无论(x,y)是多少，(xx,yy)都是(x,y)所在的小九格中的左下角，从(xx,yy)出发，遍历小九格就比较容易了。
3. 直接在原数组中记录答案，然后递归求解，如果解答失败，就把记录的该答案清空。这种方法在递归中经常用到，好处是不需要占用额外的内存，也不会产生”脏数据”干扰。
4. 递归方法solveSudoku(char[][] board, int i)输入的位置i一定是没填的位置，查找下一个没填的位置k用这句话就行了while(++k < 81 && board[k / 9][k % 9] != '.') ;

算法如下，本算法击败了92%的java解法。

    public static void solveSudoku(char[][] board) {
        int k = -1;
        while(++k < 81 && board[k / 9][k % 9] != '.') ;
        if(k < 81) solveSudoku(board, k);
    }
    
    public static boolean solveSudoku(char[][] board, int i) {
        int y = i / 9;
        int x = i % 9;
        for(char c = '1'; c <= '9'; c++) {
            if(check(board, i, c)) {
                board[y][x] = c;
                int k = i;
                while(++k < 81 && board[k / 9][k % 9] != '.') ;
                if(k == 81) return true;
                if(solveSudoku(board, k)) return true;
                board[y][x] = '.';
            }
        }
        return false;
    }
    
    public static boolean check(char[][] board, int index, char c) {
        int y = index / 9;
        int x = index % 9;
        int xx = x / 3 * 3, yy = y / 3 * 3;
        for(int i = 0; i < 9; i++) {
            if(board[y][i] == c) return false;
            if(board[i][x] == c) return false;
            if(board[yy + i / 3][xx + i % 3] == c) return false;
        }
        return true;
    }