37. 解数独
题目描述: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
backTrack
方法:
- 这是一个递归方法,用于回溯填充数独。它使用两层嵌套循环来遍历数独的每个格子。
- 如果当前格子为空(即
board[i][j] == '.'
),则尝试填充数字 '1' 到 '9'。 - 对于每个尝试填充的数字,首先调用
Judge
方法来判断当前数字是否符合数独的规则。 - 如果当前数字符合数独的规则,则将其填充到当前格子,并递归调用
backTrack
方法填充下一个格子。 - 如果递归调用返回
true
,表示数独已经填充完成,则直接返回true
,结束递归。 - 如果递归调用返回
false
,表示当前填充的数字导致后续无法填充完成,则回溯到上一步,尝试下一个数字。 - 如果所有数字都尝试过了仍然无法填充完成,则返回
false
。
Judge
方法:
- 这是一个辅助方法,用于判断当前填充的数字是否符合数独的规则。
- 它通过检查当前行、当前列和当前小九宫格来确定当前数字是否与已有的数字冲突。
- 如果存在冲突,则返回
false
,否则返回true
。
class Solution {
public void solveSudoku(char[][] board) {
backTrack(board);
}
public boolean backTrack(char[][] board) {
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
if (board[i][j] != '.') continue;
for (char k = '1'; k <= '9'; k++) {
if(!Judge(board, i, j, k))continue;
board[i][j] = k;
if (backTrack(board)) return true;
board[i][j] = '.';
}
return false;
}
}
return true;
}
public boolean Judge(char[][] board, int x, int y, char k) {
for (int i = 0; i < 9; i++) if (board[x][i] == k) return false;
for (int i = 0; i < 9; i++) if (board[i][y] == k) return false;
int nx = (x / 3) * 3;
int ny = (y / 3) * 3;
for (int i = nx; i < nx + 3; i++) {
for (int j = ny; j < ny + 3; j++) {
if (board[i][j] == k) return false;
}
}
return true;
}
}