36. Valid Sudoku(合法數獨)
Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.
The Sudoku board could be partially filled, where empty cells are filled with the character '.'.
A partially filled sudoku which is valid.
Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.
關於數獨的簡介：
There are just 3 rules to Sudoku.
1.Each row must have the numbers 1-9 occuring just once.
2.Each column must have the numbers 1-9 occuring just once.
3.And the numbers 1-9 must occur just once in each of the 9 sub-boxes of the grid.
題目大意：
判斷一個給定的二維數組是否是一個合法的數獨矩陣。
思路:
採用set這一容器，來進行去重。
1.判斷每一行是否合法。
2.判斷每一列是否合法。
3.判斷每一個九宮格是否合法。
代碼如下：
class Solution {
public:
bool isValidSudoku(vector<vector<char>>& board)
{
set<char> mySet;
//1.判斷每一行是否合法
for (int row = 0; row < 9; row++)
{
//cout<<"檢測行:"<<row<<endl;
for (int column = 0; column < 9; column++)
{
if (board[row][column] == '.')
{
continue;
}
if (mySet.find(board[row][column]) == mySet.end())
{
mySet.insert(board[row][column]);
}
else
{
return false;
}
}
mySet.clear();
}
//2.判斷每一列是否合法
for (int row = 0; row < 9; row++)
{
//cout<<"檢測列:"<<row<<endl;
for (int column = 0; column < 9; column++)
{
if (board[column][row] == '.')
{
continue;
}
if (mySet.find(board[column][row]) == mySet.end())
{
mySet.insert(board[column][row]);
}
else
{
return false;
}
}
mySet.clear();
}
//3.判斷每一個九宮格是否合法
for (int row = 0; row < 9; row += 3)
{
for (int column = 0; column < 9; column += 3)
{
for (int i = row; i < row + 3; i++)
{
for (int j = column; j < column + 3; j++)
{
if (board[i][j] == '.')
{
continue;
}
if (mySet.find(board[i][j]) == mySet.end())
{
mySet.insert(board[i][j]);
}
else
{
return false;
}
}
}
mySet.clear();
}
}
return true;
}
};2016-08-13 12:21:54