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
public boolean solveSudoku(char[][] board) {
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
如果为空格就依次把1-9填入
if (board[i][j] == '.') {
for (char k = '1'; k <= '9'; k++) {
// 判断填入的数字是否合法
if (isValid(i, j, k, board)) {
board[i][j] = k;
// 递归
boolean result = solveSudoku(board);
if(result == true)
return true;
board[i][j]='.'; //不合法就回溯,回溯前先重新赋值为空
}
}
// 填写完成后返回
return false;
}
}
}
return true;
}
/**
* 判断放入棋盘的数字是否合法
*/
public boolean isValid(int row, int col, char val, char[][] board) {
// 验证一行
for (int i = 0; i < 9; i++) { // 判断行里是否重复
if (board[row][i] == val) {
return false;
}
}
// 验证一列
for (int j = 0; j < 9; j++) { // 判断行里是否重复
if (board[j][col] == val) {
return false;
}
}
// 验证3*3
int startRow = (row / 3) * 3;
int startCol = (col / 3) * 3;
for (int i = startRow; i < startRow + 3; i++) { // 判断9方格里是否重复
for (int j = startCol; j < startCol + 3; j++) {
if (board[i][j] == val ) {
return false;
}
}
}
return true;
}