编写一个程序，通过填充空格来解决数独问题。
数独的解法需 遵循如下规则：
数字1-9在每一行只能出现一次。
数字1-9在每一列只能出现一次。
数字1-9在每一个以粗实线分隔的3x3宫内只能出现一次。（请参考示例图）
数独部分空格内已填入了数字，空白格用’.’表示。
示例 1：
输入：board = [[“5”,”3”,”.”,”.”,”7”,”.”,”.”,”.”,”.”],[“6”,”.”,”.”,”1”,”9”,”5”,”.”,”.”,”.”],[“.”,”9”,”8”,”.”,”.”,”.”,”.”,”6”,”.”],[“8”,”.”,”.”,”.”,”6”,”.”,”.”,”.”,”3”],[“4”,”.”,”.”,”8”,”.”,”3”,”.”,”.”,”1”],[“7”,”.”,”.”,”.”,”2”,”.”,”.”,”.”,”6”],[“.”,”6”,”.”,”.”,”.”,”.”,”2”,”8”,”.”],[“.”,”.”,”.”,”4”,”1”,”9”,”.”,”.”,”5”],[“.”,”.”,”.”,”.”,”8”,”.”,”.”,”7”,”9”]]
输出：[[“5”,”3”,”4”,”6”,”7”,”8”,”9”,”1”,”2”],[“6”,”7”,”2”,”1”,”9”,”5”,”3”,”4”,”8”],[“1”,”9”,”8”,”3”,”4”,”2”,”5”,”6”,”7”],[“8”,”5”,”9”,”7”,”6”,”1”,”4”,”2”,”3”],[“4”,”2”,”6”,”8”,”5”,”3”,”7”,”9”,”1”],[“7”,”1”,”3”,”9”,”2”,”4”,”8”,”5”,”6”],[“9”,”6”,”1”,”5”,”3”,”7”,”2”,”8”,”4”],[“2”,”8”,”7”,”4”,”1”,”9”,”6”,”3”,”5”],[“3”,”4”,”5”,”2”,”8”,”6”,”1”,”7”,”9”]]
解释：输入的数独如上图所示，唯一有效的解决方案如下所示：
提示：
- board.length == 9
- board[i].length == 9
- board[i][j] 是一位数字或者 ‘.’
- 题目数据 保证 输入数独仅有一个解
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| /**
* 37. 解数独
*/
public class LeetCode37 {
/**
* 数独问题
* 根据题意：
* 1、数独一般都是9*9的数字棋盘等
* 2、数独同一个数字在同一行，或者同一列，或者在相同的3*3宫格里不能重复，只能填写0~9的数字
* 解法
* 1、也只能用暴力破解，使用两层for循环确定当前的坐标，再使用一层0~9的循环回溯判断当前数字是否准确
* 2、看到题目的返回类型是空，所以结果应该在原输入二维数组上修改
*/
public void solveSudoku(char[][] board) {
solveSudokuHelper(board);
}
private boolean solveSudokuHelper(char[][] board) {
for (int i = 0; i < board.length; i++) {
for (int j = 0; j< board[0].length; j++) {
//当前坐标值
char cru = board[i][j];
if (cru != '.') {//当前坐标已经有数字，跳过
continue;
}
for (char k = '1'; k <= '9'; k++) {//1~9的数字当前坐标试错
if (isValue(i, j, k, board)) {
board[i][j] = k;
if (solveSudokuHelper(board)) {//找到结果，直接返回，这里是全文最最关键的点，要细致的品
return true;
}
board[i][j] = '.';//试错失败回溯
}
}
//目前上述1~9都试错没找到，那么就返回false,这个点不能漏，很关键
return false;
}
}
return true;
}
/**
* 判断当前数字是否合法
*/
private boolean isValue(int row, int col, char val, char[][] board) {
//在同一行没有重复,这里必须要是从0到9开始遍历
for (int j = 0; j < 9; j++) {
if (board[row][j] == val) {
return false;
}
}
//在同一列没有重复，这里必须要是从0到9开始遍历 重点
for (int i = 0; i < 9; i++) {
if (board[i][col] == val) {
return false;
}
}
int startRow = row / 3 * 3;
int stratCol = col / 3 * 3;
for (int i = startRow; i < startRow + 3; i++) {
for (int j = stratCol; j < stratCol + 3; j++) {
if (board[i][j] == val) {
return false;
}
}
}
return true;
}
}
|