[LeetCode]037. 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.

Solution: use the back tracking and brute force.

Note:

1) isValid()用来判断board[i][j]插入的字符是否合法;

2) char类型的变量可以直接比较大小并自增；

3) 填充数字时，采用int pos来记录位置，int i = pos/9, int j = pos%9, i,j 表示行列的位置。

4) 进行回溯的时候必须有返回值，以判定是否合法。

5) 代码48~50行一开始写成

if(isValid(board,i,j)){
	solveS(board,pos+1);
}

没有使用isValid(board,i,j) && solveS(board,pos+1)，导致了递归方法没有办法回溯，只能继续递归下去直到退出。

这样默认为上一级填充的字符均为正确的，会导致最终结果有空缺的'.' 。

public class Solution {
    public boolean isValid(char[][] board, int x, int y){
        Set<Character> rowCheck = new HashSet<Character>();
        for(int i=0; i<9; i++){
            if(rowCheck.contains(board[x][i])){
                return false;
            }
            if(board[x][i]<='9' && board[x][i]>='1'){
                rowCheck.add(board[x][i]);
            }
        }
        
        Set<Character> colCheck = new HashSet<Character>();
        for(int i=0; i<9; i++){
            if(colCheck.contains(board[i][y])){
                return false;
            }
            if(board[i][y]<='9' && board[i][y]>='1'){
                colCheck.add(board[i][y]);
            }
        }
        
        Set<Character> squCheck = new HashSet<Character>();
        for(int i=0; i<3; i++){
            for(int j=0; j<3; j++){
                int m =x/3*3+i;
                int n = y/3*3+j;
                if(squCheck.contains(board[m][n])){
                    return false;
                }
                if(board[m][n]>='1' && board[m][n]<='9'){
                    squCheck.add(board[m][n]);
                }
            }
        }
        return true;
    }
    
    public boolean solveS(char[][] board, int pos){
        if(pos > 9*9-1){
            return true;
        }
        int i = pos/9;
        int j = pos%9;
        if(board[i][j] == '.'){
            for(char k='1'; k<='9'; k++){
                board[i][j] = k;
                if(isValid(board,i,j) && solveS(board,pos+1)){
                    //must use '&&' and 'return' here, for back tracking
                    return true;
                }
                board[i][j] = '.';
            }
        }else{
            return solveS(board, pos+1);
        }
        return false;
    }
    
    public void solveSudoku(char[][] board) {
        // Start typing your Java solution below
        // DO NOT write main() function
        solveS(board, 0);
    }
}