1.Description
编写一个程序,通过填充空格来解决数独问题。
一个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。
空白格用 ‘.’ 表示。
2.Example
Input: 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"]]
Output: [["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"]]
3.Solution
定义line[i][digit],column[j][digit],block[i/3][j/3][digit]来分别表示数独矩阵中第i行digit这个数字是否已经出现,第j列digit这个数字是否已出现,在图中的九个九宫格中第i/3行j/3列的这个九宫格中digit这个数字是否已经出现。
定义spaces来储存需要填数的行数和列数。
先将题中给出的已经存在的数在line,column,block中定义为true,表示已经出现过。
然后运用dfs:判断一个digit如果在line,column,block中都没有出现过,就填到矩阵中去,然后进行下一个递归。最后当所有数都顺利填完了,或者递归过程中遇到一个空找不到合适的数填进去了(在行,列,九宫格中都出现过了),就回溯到上一层。注意回溯回来的时候要把line,column,block都改回false,表示没有填成功。
public class Solution {
private boolean[][] line = new boolean[9][9];
private boolean[][] column = new boolean[9][9];
private boolean[][][] block = new boolean[3][3][9];
private List<int[]> spaces = new ArrayList<int[]>();
private boolean valid = false;
public void solveSudoku(char[][] board) {
for(int i=0;i<9;i++) {
for(int j=0;j<9;j++) {
if(board[i][j]=='.') {
spaces.add(new int[] {
i,j});
}else {
int digit = board[i][j]-'0'-1;
line[i][digit]=column[j][digit]=block[i/3][j/3][digit]=true;
}
}
}
dfs(board,0);
}
public void dfs(char[][] board,int pos) {
if(pos==spaces.size()) {
valid = true;
return;
}
int i = spaces.get(pos)[0];
int j = spaces.get(pos)[1];
for(int digit=0;digit<9&&!valid;digit++) {
if(!line[i][digit]&&!column[j][digit]&&!block[i/3][j/3][digit]) {
line[i][digit]=column[j][digit]=block[i/3][j/3][digit]=true;
board[i][j] = (char) (digit+'0'+1);
dfs(board, pos+1);
line[i][digit]=column[j][digit]=block[i/3][j/3][digit]=false;
}
}
}
}