LeetCode-036-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。

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

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/valid-sudoku

解题思路

HashMap 的空间消耗大, 采用布尔二维数组
boolean[][] rows = new boolean[9][9]
boolean[][] cols = new boolean[9][9]
boolean[][] boxes = new boolean[9][9]
rows[i][num - 1]代表第 i 行的数 num 是否出现过, 是则为 true

因为数组下标为0到长度 - 1, 所以用 num - 1 代表num

遍历9 * 9的数独时, 依次判断遍历的数字在该行该列该九宫格是否出现过, 出现过立刻返回 false, 没出现过则设为出现过 true
遍历完之前没返回结果则说明符合有效数独

代码

class Solution {
    public boolean isValidSudoku(char[][] board) {
        boolean[][] rows = new boolean[9][9];
        boolean[][] cols = new boolean[9][9];
        boolean[][] boxes = new boolean[9][9];
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                char numChar = board[i][j];
                if (numChar != '.') {
                    int num = numChar - '0';
                    if (rows[i][num - 1]) {
                        return false;
                    } else {
                        rows[i][num - 1] = true;
                    }
                    if (cols[j][num - 1]) {
                        return false;
                    } else {
                        cols[j][num - 1] = true;
                    }
                    if (boxes[i / 3 * 3 + j / 3][num - 1]) {
                        return false;
                    } else {
                        boxes[i / 3 * 3 + j / 3][num - 1] = true;
                    }
                }
            }
        }
        return true;
    }
}