LeetCode刷题笔录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.

和N-Queens那题想法差不多，对于每个格子，从1到9都填进去一遍试一下，然后check是否和当前已有的元素冲突。如果不冲突，则递归进行下一个格子。可以按行前进一格也可以按列前进一格。注意尝试填进去的数字再递归出来之后要改回'.‘。

public class Solution {
    public void solveSudoku(char[][] board) {
        solveSudoku(board, 0, 0);
    }
    
    public boolean solveSudoku(char[][] board, int row, int col){
        if(col >= 9)
            return solveSudoku(board, row + 1, 0);
        if(row == 9)
            return true;
        if(board[row][col] == '.'){
            for(char i = '1'; i <= '9'; i++){
                board[row][col] = i;
                if(isValid(board, row, col, i)){
                    if(solveSudoku(board, row, col + 1))
                        return true;
                }
                board[row][col] = '.';
            }
        }
        else{
            return solveSudoku(board, row, col + 1);
        }
        return false;
    }
    public boolean isValid(char[][] board, int row, int col, char value){
        //check if duplicate exists in the same row or column
        for(int i = 0; i < 9; i++){
            if(i != col && value == board[row][i])
                return false;
            if(i != row && value == board[i][col])
                return false;
        }
        
        //check if duplicate exists in the same cube
        int cubeRowStart = row / 3 * 3;
        int cubeColStart = col / 3 * 3;
        for(int i = cubeRowStart; i < cubeRowStart + 3; i++){
            for(int j = cubeColStart; j < cubeColStart + 3; j++){
                if(i != row && value == board[i][j])
                    return false;
                if(j != col && value == board[i][j]) 
                    return false;
            }
        }
        
        return true;
    }
}