36. 有效的数独
题目描述
判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。
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。
上图是一个部分填充的有效的数独。数独部分空格内已填入了数字，空白格用 '.' 表示。
示例 1:
输入:
[
["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"]
]
输出: true
示例 2:
输入:
[
["8","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"]
]
输出: false
解释: 除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。
但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。
说明:
一个有效的数独（部分已被填充）不一定是可解的。只需要根据以上规则，验证已经填入的数字是否有效即可。给定数独序列只包含数字 1-9 和字符 '.' 。给定数独永远是 9x9 形式的。
玩家需要根据9×9盘面上的已知数字，推理出所有剩余空格的数字，并满足每一行、每一列、每一个粗线宫（3*3）内的数字均含1-9，不重复。数独盘面是个九宫，每一宫又分为九个小格。在这八十一格中给出一定的已知数字和解题条件，利用逻辑和推理，在其他的空格上填入1-9的数字。使1-9每个数字在每一行、每一列和每一宫中都只出现一次，所以又称“九宫格”。
方法1
递归解决方案
class Solution {
public:
bool isValidSudoku(vector<vector<char> > &board) {
if(board.size()!=9 || board[0].size()!=9)
return false;
return isValidSudokuCore(board,0);
}
bool isValidSudokuCore(vector<vector<char> > &board,int index)
{
if(index == 81)
return true;
int row = index/9;
int col = index%9;
if(board[row][col] != '.')
{
if(isvalid(board,row,col))
return isValidSudokuCore(board,index+1);
else
return false;
}
else
return isValidSudokuCore(board,index+1);
//这条语句不需要，上面的语句一定能返回结果
//return false;
}
bool isvalid(vector<vector<char> > &board,int row,int col)
{
if(board[row][col]-'0'>9 || board[row][col]-'0'<1)
return false;
for(int i = 0; i < 9; i++)
{
if(i != col && board[row][col] == board[row][i])
return false;
if(i != row && board[row][col] == board[i][col])
return false;
}
for(int i = (row/3)*3; i < (row/3)*3+3;i++)
{
for(int j = (col/3)*3; j < (col/3)*3+3; j++)
{
if(!(i == row && j == col) && board[row][col] == board[i][j])
return false;
}
}
return true;
}
};
方法2
上面方案的循环实现
class Solution {
public:
bool isValidSudoku(vector<vector<char> > &board)
{
if(board.size()!=9 || board[0].size()!=9)
return false;
for(int row = 0; row < 9; row++)
{
for(int col = 0; col < 9; col++)
{
if(board[row][col] == '.')
continue;
else
{
if(board[row][col]-'0'>9 || board[row][col]-'0'<1)
return false;
for(int i = 0; i < 9; i++)
{
if(i != col && board[row][col] == board[row][i])
return false;
if(i != row && board[row][col] == board[i][col])
return false;
}
for(int i = (row/3)*3; i < (row/3)*3+3;i++)
{
for(int j = (col/3)*3; j < (col/3)*3+3; j++)
{
if(!(i == row && j == col) && board[row][col] == board[i][j])
return false;
}
}
}
}
}
return true;
}
};
方法三
利用辅助数组
class Solution {
public:
bool isValidSudoku(vector<vector<char> > &board) {
////第二维下标表示这个数在当前范围是否已经出现过
int rows[9][9]={0},cols[9][9]={0},cube[9][9]={0};
for(int i=0;i<9;i++)
{
for(int j=0;j<9;j++)
{
char num = board[i][j];
if(num != '.')
{
if(num-'0'>9 || num-'0'<1)
return false;
if(rows[i][num-'1'] == 1) //判断当前行
return false;
else
rows[i][num-'1'] = 1;
if(cols[j][num-'1'] == 1) //判断当前列
return false;
else
cols[j][num-'1'] = 1;
//这里的第一个3表示横着看有3个列块
if(cube[3*(i/3)+j/3][num-'1'] == 1) //判断当前块
return false;
else
cube[3*(i/3)+j/3][num-'1'] = 1;
}
}
}
return true;
}
};
