LeetCode-探索-初级-数组-有效的数独-java
有效的数独
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。
我的方法：
还是比较简单的题，方法就是行检查，列检查，小正方形检查
分为三种分开的方法：
第一种：行检查，列检查，正方形检查
第二种：行列检查，正方形检查
第三种：三个代码放在一个中检查
反正并没有什么不同。（考虑到BitSet初始化的时间的话，看似后面两种比较复杂，但也是非常得不偿失的）
方便起见，我将所有的代码一起粘上
public boolean isValidSudoku(char[][] board) {
//check rows
BitSet bitSet = new BitSet(9);
for (int i = 0 ; i < 9; i ++) {
bitSet.clear(); //for each row, bitSet should be clean
for (int j = 0 ; j < 9 ; j ++)
if (!check(board, i, j, bitSet))
return false;
}
//check cols
for (int col = 0 ; col < 9 ; col ++) {
bitSet.clear();
for (int row = 0 ; row < 9 ; row ++)
if (!check(board, row, col, bitSet))
return false;
}
//check small square
//totally 3 rows and 3 cols small squares
//each small square has 3 rows and 3 cols
/**
* squareRow 0 --- board[0~2][]
* squareRow 1 --- board[3~5][]
* squareRow 2 --- board[6~8][]
* squareRow n --- board[ 3*n+row ][]
* ...
* squareCol m --- board[ 3*n+row ][ 3*m+col ]
*/
for (int squareRow = 0 ; squareRow < 3 ; squareRow ++) {
for (int squareCol = 0 ; squareCol < 3 ; squareCol ++) {
bitSet.clear();
for (int row = 0 ; row < 3 ; row ++)
for (int col = 0 ; col < 3 ; col ++) {
if (!check(board, squareRow * 3 + row, 3 * squareCol + col, bitSet))
return false;
}
}
}
return true;
}
public boolean isValidSudoku1(char[][] board) {
//check rows and cols
BitSet[] bitSets = new BitSet[]{
new BitSet(9), new BitSet(9), new BitSet(9),
new BitSet(9), new BitSet(9), new BitSet(9),
new BitSet(9), new BitSet(9), new BitSet(9)
}; //cols
BitSet bitSet = new BitSet(9); //row
for (int i = 0 ; i < 9; i ++) {
bitSet.clear(); //for each row, bitSet should be clean
for (int j = 0 ; j < 9 ; j ++)
if (board[i][j] != '.')
if (!check1(board, i, j, bitSet) || !check1(board, i, j, bitSets[j]))
return false;
}
//check small square
//totally 3 rows and 3 cols small squares
//each small square has 3 rows and 3 cols
/**
* squareRow 0 --- board[0~2][]
* squareRow 1 --- board[3~5][]
* squareRow 2 --- board[6~8][]
* squareRow n --- board[ 3*n+row ][]
* ...
* squareCol m --- board[ 3*n+row ][ 3*m+col ]
*/
for (int squareRow = 0 ; squareRow < 3 ; squareRow ++) {
for (int squareCol = 0 ; squareCol < 3 ; squareCol ++) {
bitSet.clear();
for (int row = 0 ; row < 3 ; row ++)
for (int col = 0 ; col < 3 ; col ++) {
if (!check(board, squareRow * 3 + row, 3 * squareCol + col, bitSet))
return false;
}
}
}
return true;
}
public boolean isValidSudoku10(char[][] board) {
//check rows and cols
BitSet[] rowBitSets = new BitSet[]{
new BitSet(9), new BitSet(9), new BitSet(9),
new BitSet(9), new BitSet(9), new BitSet(9),
new BitSet(9), new BitSet(9), new BitSet(9)
}; //cols
BitSet bitSet = new BitSet(9); //squares
BitSet[] colBitSets = new BitSet[]{
new BitSet(9), new BitSet(9), new BitSet(9),
new BitSet(9), new BitSet(9), new BitSet(9),
new BitSet(9), new BitSet(9), new BitSet(9)
};
//check small square
//totally 3 rows and 3 cols small squares
//each small square has 3 rows and 3 cols
/**
* squareRow 0 --- board[0~2][]
* squareRow 1 --- board[3~5][]
* squareRow 2 --- board[6~8][]
* squareRow n --- board[ 3*n+row ][]
* ...
* squareCol m --- board[ 3*n+row ][ 3*m+col ]
*/
int indexRow = 0;
int indexCol = 0;
for (int squareRow = 0 ; squareRow < 3 ; squareRow ++) {
for (int squareCol = 0 ; squareCol < 3 ; squareCol ++) {
bitSet.clear();
for (int row = 0 ; row < 3 ; row ++)
for (int col = 0 ; col < 3 ; col ++) {
indexRow = squareRow * 3 + row;
indexCol = squareCol * 3 + col;
if (board[indexRow][indexCol] != '.')
if (!check1(board, indexRow, indexCol, bitSet) ||
!check1(board, indexRow, indexCol, rowBitSets[indexRow])||
!check1(board, indexRow, indexCol, colBitSets[indexCol]))
return false;
}
}
}
return true;
}
public boolean check(char[][] board, int row, int col, BitSet bitSet) {
if (board[row][col] != '.')
if (bitSet.get(board[row][col])) //if it exists
return false;
else
bitSet.set(board[row][col]);
//end if
//end if
return true;
}
public boolean check1(char[][] board, int row, int col, BitSet bitSet) {
if (bitSet.get(board[row][col])) //if it exists
return false;
else
bitSet.set(board[row][col]);
//end if
//end if
return true;
}官网上给的最佳方法（emmm总是要有这一步的~）：
先确定行列，比较同一行之后的数；比较同一列之后的数；之后在所在的九宫格中，比较之后的列的数（虽然答案上很容易看出来是行，但是，行列是等价的）
上代码还是比较容易一点（其实画一个图更容易）：
/**
* method:
* 1.Identify the row count and line count.
* 2.Compare the following numbers in current line.
* 3.Compare the following numbers in current row.
* 4.Compare the following numbers in following lines that in current square.
* @param board
* @return
*/
public boolean isValidSudoku11(char[][] board) {
//step 1 in comment
for (int rowCount = 0 ; rowCount < 9 ; rowCount ++)
for (int lineCount = 0 ; lineCount < 9 ; lineCount ++) {
if (board[rowCount][lineCount] == '.')
continue;
//step 2 in comment
for (int row = rowCount + 1 ; row < 9 ; row ++)
if (board[rowCount][lineCount] == board[row][lineCount])
return false;
//step 3 in comment
for (int line = lineCount + 1 ; line < 9 ; line ++)
if (board[rowCount][lineCount] == board[rowCount][line])
return false;
//step 4 in comment
/**
* For this step:
* If we want to select a sub array, we must know the range of row and col
* The range of col is obviously lineCount+1 to lineCount%3==0
* The range of row is obviously 3(from rowCount/3*3 to rowCount/3*3+3)
*/
int rowLimitation = rowCount / 3 * 3 + 3;
for (int col = lineCount + 1 ; col % 3 != 0 ; col ++)
for (int row = rowCount / 3 * 3 ; row < rowLimitation ; row ++)
if (board[rowCount][lineCount] == board[row][col])
return false;
}
//end loop
return true;
}其实挺奇怪的，我修改之后的代码也不能排在时间消耗最少的里面