夯实算法-有效的数独

题目：有效的数独

请你判断一个 9 x 9 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。

1. 数字 1-9 在每一行只能出现一次。
2. 数字 1-9 在每一列只能出现一次。
3. 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。（请参考示例图）

注意：

• 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。
• 空白格用 '.' 表示。

示例 1：

输入： board = 
[["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： board = 
[["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。

提示：

• board.length == 9
• board[i].length == 9
• board[i][j] 是一位数字（1-9）或者 '.'

解题思路

数组三条规则 1、同一行无重复 2、同一列无重复 3、同一个矩阵无重复 1、2由hl数组界定 3由jz数组界定

如何判断是否在同一个子矩阵中呢? 找规律发现若i1/3 == i2/3 j1/3 == j2/3,就说明在同一个子矩阵中。 因此构建的是[3][3]的数组来区别子矩阵。 当然因为i/3 和 j/3 都在[0,2]之内，因此可以用一个函数 area = i/3*3+j/3来界定，area属于[0,9) 该方法完结！

代码实现

public boolean isValidSudoku(char[][] board) {
    boolean[][][] hl = new boolean[2][9][9];
    //第一个0代表行 1代表列  第二个9是9行/9列  第三个9代表9个数字
    boolean[][][] jz = new boolean[3][3][9]; //i j代表第i行第j列个3x3矩阵
    for (int i = 0; i < 2; i++)
        for (int j = 0; j < 9; j++)
            for (int k = 0; k < 9; k++)
                hl[i][j][k] = false;
    int numberTemp = 2;
    for (int i = 0; i < 9; i++) {
        for (int j = 0; j < 9; j++) {
            if (board[i][j] == '.')
                continue;
            numberTemp = (board[i][j] - '0') - 1; //numberTemp 0~8  对应1~9
            //System.out.println(i+" "+j+" "+numberTemp);

            if (hl[0][i][numberTemp] == true || hl[1][j][numberTemp] == true) {
                //System.out.println(i+"!!! "+j+" "+numberTemp);
                return false;
            } else {
                hl[0][i][numberTemp] = true;
                hl[1][j][numberTemp] = true;
            } //处理行列

            int hTemp = i / 3;
            int lTemp = j / 3;
            if (jz[hTemp][lTemp][numberTemp] == true) {
                return false;
            } else {
                jz[hTemp][lTemp][numberTemp] = true;
            }
        }
    }
    return true;
}

运行结果

复杂度分析

• 空间复杂度：O(n)
• 时间复杂度：O(n^2)

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