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。9*9
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。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/valid-sudoku
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从上层往下一层一层判断，难点在于判断当前数字是否合法
class Solution {
private int targetN = 9;// 大宫格的n
private int blockN = 3; // 小宫格的n
private boolean[][] columnNumExist; // 列序号-数字:是否存在
private boolean[][] rowNumExist; // 行序号-数字：是否存在
private boolean[][][] blockNumExist; // 第几个3*3宫-数字：是否存在
/**
* 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。
* @param board
* @return
*/
public boolean isValidSudoku(char[][] board) {
columnNumExist = new boolean[targetN][targetN];
rowNumExist = new boolean[targetN][targetN];
blockNumExist = new boolean[blockN][blockN][targetN];
for (int i = 0; i < targetN; i++) {
for (int j = 0; j < targetN; j++) {
// '.'不需要进行判断
if ('.' == board[i][j]) {
continue;
}
// board[i][j] - '0' - 1 => 将字符转化为数字1-9，然后再-1，存储数字0-8
if (!curValid(i, j, board[i][j] - '0' - 1)) {
return false;
}
recordNum(i, j, board[i][j] - '0' - 1);
}
}
return true;
}
/**
* 判断当前值在这个坐标上是否满足
* @param row 行
* @param column 列
* @param num 数字
* @return
*/
private boolean curValid(int row, int column, int num) {
// 行是否已经包括
if (rowNumExist[row][num]) {
return false;
}
// 列是否已经包括
if (columnNumExist[column][num]) {
return false;
}
// 3*3宫格是否已经包括
if (blockNumExist[row / blockN][column / blockN][num]) {
return false;
}
return true;
}
/**
* 记录当前坐标的值
* @param row 行
* @param column 列
* @param num 数字
* @return
*/
private void recordNum(int row, int column, int num) {
rowNumExist[row][num] = true;
columnNumExist[column][num] = true;
blockNumExist[row / blockN][column / blockN][num] = true;
}
}
运行效率
进一步优化，用二进制数替代二位数组，二级制的位运算效率很高
class Solution {
private int targetN = 9;// 大宫格的n
private int blockN = 3; // 小宫格的n
private int curRowBit; // 用一个数字表示当前行的情况，相当于 010100000 9个比特位表示，1表示已经选过
private int[] columnBit;
private int[] blockBit;
/**
* 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。
* @param board
* @return
*/
public boolean isValidSudoku(char[][] board) {
columnBit = new int[targetN];
blockBit = new int[targetN];
for (int i = 0; i < targetN; i++) {
curRowBit = 0; // 存储当前行的bit位数
for (int j = 0; j < targetN; j++) {
// '.'不需要进行判断
if ('.' == board[i][j]) {
continue;
}
// board[i][j] - '0' - 1 => 将字符转化为数字1-9，然后再-1，存储数字0-8
if (!curValid(i, j, board[i][j] - '1')) {
return false;
}
recordNum(i, j, board[i][j] - '1');
}
}
return true;
}
/**
* 判断当前值在这个坐标上是否满足
* @param row 行
* @param column 列
* @param num 数字
* @return
*/
private boolean curValid(int row, int column, int num) {
// 行是否已经包括
if ((curRowBit >> num) % 2 == 1) {
return false;
}
// 列是否已经包括
if ((columnBit[column] >> num) % 2 == 1) {
return false;
}
// 3*3宫格是否已经包括
int cnt = row / blockN * blockN + column / blockN;
if ((blockBit[cnt] >> num) % 2 == 1) {
return false;
}
return true;
}
/**
* 记录当前坐标的值
* @param row 行
* @param column 列
* @param num 数字
* @return
*/
private void recordNum(int row, int column, int num) {
curRowBit += 1 << num;
columnBit[column] += 1 << num;
int cnt = row / blockN * blockN + column / blockN;
blockBit[cnt] += 1 << num;
}
}
二进制数替代二位数组
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