36. Valid Sudoku(合法数独)
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.
关于数独的简介：
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.
题目大意：
判断一个给定的二维数组是否是一个合法的数独矩阵。
思路:
采用set这一容器，来进行去重。
1.判断每一行是否合法。
2.判断每一列是否合法。
3.判断每一个九宫格是否合法。
代码如下：
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class
Solution {
public
:
bool
isValidSudoku(vector<vector<
char
>>& board)
{
set<
char
> mySet;
//1.判断每一行是否合法
for
(
int
row = 0; row < 9; row++)
{
//cout<<"检测行:"<<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.判断每一列是否合法
for
(
int
row = 0; row < 9; row++)
{
//cout<<"检测列:"<<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.判断每一个九宫格是否合法
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
;
}
};
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本文转自313119992 51CTO博客，原文链接:http://blog.51cto.com/qiaopeng688/1837537
