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题目描述(引自LeetCode)

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。

题目分析

这是一道判断9x9数独是否有效(而非是否可解)的题目，只需要按照题目给定的3个判断条件逐一判断即可。为了减少遍历次数，需要在针对board[][]的一次遍历中完成三个条件的判断，详见代码及注释。

代码实现

class Solution {public boolean isValidSudoku(char[][] board) {// 初始化行数组，列数组，宫数组boolean[][] rows = new boolean[9][9];boolean[][] columns = new boolean[9][9];boolean[][] boxes = new boolean[9][9];for (int i = 0; i < 9; i++) {for (int j = 0; j < 9; j++) {// 非空白格if (board[i][j] != '.') {// char转intint num = board[i][j] - '1';// 用于枚举子数独int k = (i / 3) * 3 + j / 3;// 判断是否符合要求if (rows[i][num] || columns[j][num] || boxes[k][num]) return false;// 更新行数组，列数组，宫数组rows[i][num] = true;columns[j][num] = true;boxes[k][num] = true;}}}// 返回结果return true;}}

代码分析

对代码进行分析，算法仅仅是对数独9宫格进行了一次遍历，固定为81次访问，因此，时间复杂度为O(1)，而就空间而言，仅仅使用了常数级别的额外空间，因此，空间复杂度为O(1)。

执行结果

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