[LeetCode37]Soduko Solver_valid soduko-优快云博客

本文介绍了一种通过编程解决数独谜题的方法。利用回溯算法填充空格，并验证每一步是否符合数独规则，确保解决方案的唯一性和正确性。提供了Java和C++两种语言的实现代码。

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.

The idea is straight forward, put the number in empty place. and check if it is valid and solverable.

Java

public boolean solveSudoku(char[][] board) {
		for(int i=0;i<9;i++){
			for(int j = 0;j<9;j++){
				if(board[i][j]=='.'){
					for(int k=1;k<=9;k++){
						board[i][j]=(char) (k+48);
						if(validSudoku(board, i, j) && solveSudoku(board) )
							return true;
						board[i][j] = '.';
					}
					return false;
				}
			}
		}
		return true;
    }
	public boolean validSudoku(char[][] board, int x, int y){
		for(int i=0;i<9;i++){
			if(i!=x && board[i][y]==board[x][y])
				return false;
			if(i!=y && board[x][i]==board[x][y])
				return false;
		}
		for(int i=3*(x/3);i<3*(x/3+1);i++){
			for(int j=3*(y/3);j<3*(y/3+1);j++){
				if(i!=x && j!=y && board[i][j] == board[x][y])
					return false;
			}
		}
		return true;
	}

c++

bool isValidSudoku2(vector<vector<char>> &board, int x,int y){
    int i, j;
    for(i=0; i<9; i++)
        if(i!= x && board[i][y] == board[x][y])
            return false;
    for(j = 0; j<9; j++)
        if(j!= y && board[x][j] == board[x][y])
            return false;
    for(i = 3*(x/3); i<3*(x/3+1); i++){
        for(j = 3 *(y/3); j<3*(y/3+1);j++)
            if(i!=x && j!= y && board[i][j] == board[x][y])
                return false;
    
    }
    return true;
}
bool solveSudoku(vector<vector<char> > &board) {
    for(int i=0; i<9;i++){
        for(int j=0;j<9;j++){
            if('.' == board[i][j]){
                for(int k=1;k<=9;k++){
                    board[i][j] = '0' + k;
                    if(isValidSudoku2(board,i,j) && solveSudoku(board))
                        return true;
                    board[i][j] = '.';
                }
                return false;
            }
        }
    }
    return true;
}