LeetCode36、有效的数独

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题目

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。

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。（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：

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



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 '.'

思路

这是类似数独的题目，但是除了每一行和每一列不能有重复数字外，还多了一个要求，即每一个3 * 3宫格内也不允许重复

我们可以通过循环的形式来遍历所有元素，通过三个变量分别存储行列及宫格元素，有重复返回 false，遍历完则返回 true

• 首先定义 row 存储当前行的非空元素，column 存储每一列的非空元素，box 存储每个宫格的非空元素
• 双重循环进行遍历，外层 x 遍历行，内层 y 遍历列
• 每次遍历行时，将 row 置空以存储当前行的元素，column 及 box 不需要置空
• 获取当前元素值 cur，仅当 cur 不为 . 才进行操作
• 首先进行当前行元素的判断处理，如果 row 存在当前元素，返回 false ，反之将其存储在 row
• 然后判断 column 中是否有当前列 y 的数据，无则创建列 y 并存入 cur，有则判断列 y 中是否存在 cur，存在返回 false，反之将其存入列 y
• 最后判断宫格，总共有 9 个宫格，对应下标可通过 Math.floor(x / 3) * 3 + Math.floor(y / 3) 获取
• 宫格里数据的判断同列的判断一致
• 所有元素都遍历完，则返回 true

代码实现

/**
 * @param {character[][]} board
 * @return {boolean}
 */
var isValidSudoku = function(board) {

    let row, column = new Map(), box = new Map()
    for(let x = 0; x < 9; x++) {
        row = new Map()
        for(let y = 0; y < 9; y++) {
            const cur = board[x][y]
            if(cur !== '.') {
                // 行
                if (row.has(cur)) {
                    return false
                } else {
                    row.set(cur, true)
                }

                // 列
                if (column.has(y)) {
                    if (column.get(y).has(cur)) {
                        return false
                    } else {
                        column.get(y).set(cur, true)
                    }
                } else {
                    column.set(y, new Map([[cur, true]]))
                }

                // 宫格
                const idx = Math.floor(x / 3) * 3 + Math.floor(y / 3)
                if (box.has(idx)) {
                    if (box.get(idx).has(cur)) {
                        return false
                    } else {
                        box.get(idx).set(cur, true)
                    }
                } else {
                    box.set(idx, new Map([[cur, true]]))
                }
            }
        }
    }

    return true
};