[leetcode]Sudoku Solver

Sudoku Solver

Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells are indicated by the character '.'.

You may assume that there will be only one unique solution.

A sudoku puzzle...

 

...and its solution numbers marked in red.

算法思路：

典型的DFS，与N-Queens其实没太大区别，只是冲突不同，而冲突的判断方式，在[leetcode]Valid Sudoku中已经讲得很清楚了。

代码如下：

public class Solution {
    boolean over = false;
    public void solveSudoku(char[][] board) {
        if(board == null || board.length != 9 || board[0].length != 9) return;
        boolean[][] row = new boolean[9][9];
        boolean[][] col = new boolean[9][9];
        boolean[][] matrix = new boolean[9][9];
        for(int i = 0; i < 9; i++){//初始化冲突表
            for(int j = 0; j < 9; j++){
                if(board[i][j] != '.'){
                    int n = board[i][j] - '1';
                    row[i][n] = col[j][n] = matrix[i - i % 3 + j / 3][n] = true;
                }
            }
        }
        dfs(board,0,0,row,col,matrix);
    }
    private void dfs(char[][] board,int i ,int j,boolean[][] row,boolean[][] col,boolean[][] matrix){
        if(i > 8){
            over = true;
            return;
        }
        if(board[i][j] != '.'){
            if(j < 8){
                dfs(board, i, j + 1, row, col, matrix);
            }else{
                dfs(board, i + 1, 0, row, col, matrix);
            }
        }else{
            for(int k = 0; k < 9; k++){
                if(row[i][k] || col[j][k] || matrix[i - i % 3 + j / 3][k]) continue;
                row[i][k] = col[j][k] = matrix[i - i % 3 + j / 3][k] = true;
                board[i][j] = (char)('1' + k);
                if(j < 8){
                    dfs(board, i, j + 1, row, col, matrix);
                }else{
                    dfs(board, i + 1, 0, row, col, matrix);
                }
                if(over) return;
                row[i][k] = col[j][k] = matrix[i - i % 3 + j / 3][k] = false;
                board[i][j] = '.';
            }
        }
    }
}