LeetCode 37. Sudoku Solver（解数独） DFS/hard

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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。

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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"]]

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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;
			}
		}
	}
}