回溯算法求解数独

写一个程序，解决数独求解的问题

空的格子用字符 '.'代替。

你可以假设只有一个解

一个数独 ...

...红色的数字就是解.

下面是我的代码

package com.zhazhapan.algorithm.backtracking;
public class SudokuSolve {
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		String[] string = { "..9748...", "7........", ".2.1.9...", "..7...24.", ".64.1.59.", ".98...3..", "...8.3.2.",
				"........6", "...2759.." };
		char[][] board = new char[9][9];
        // 初始化数独 
		for (int i = 0; i < 9; i++) {
			for (int j = 0; j < 9; j++) {
				board[i][j] = string[i].charAt(j);
			}
		}
		new SudokuSolve().solveSudoku(board);
	}
	public void solveSudoku(char[][] board) {
		if (board == null || board.length != 9 || board[0].length != 9) {
			return;
		}
		if (solve(board, 0, 0)) {
             // 打印结果
			for (int i = 0; i < 9; i++) {
				for (int j = 0; j < 9; j++) {
					System.out.print(board[i][j] + " ");
				}
				System.out.println();
			}
		}
	}
	public boolean solve(char[][] board, int i, int j) {
		if (j == 9) {
			if (i == 8)
				return true;
			i++;
			j = 0;
		}
		if (board[i][j] != '.') {
			return solve(board, i, j + 1);
		}
		for (char k = '1'; k <= '9'; k++) {
			if (isValid(board, i, j, k)) {
				board[i][j] = k;
				if (solve(board, i, j + 1))
					return true;
				else
					board[i][j] = '.';
			}
		}
		return false;
	}
	public boolean isValid(char[][] board, int i, int j, char c) {
		for (int k = 0; k < 9; k++) {
			if (board[i][k] != '.' && board[i][k] == c)
				return false;
			if (board[k][j] != '.' && board[k][j] == c)
				return false;
			if (board[i / 3 * 3 + k / 3][j / 3 * 3 + k % 3] != '.'
					&& board[i / 3 * 3 + k / 3][j / 3 * 3 + k % 3] == c)
				return false;
		}
		return true;
	}
}