37. 解数独 #

原题链接：LeetCode 37. 解数独

题目描述 #

编写一个程序，通过填充空格来解决数独问题。

数独的解法需** 遵循如下规则**：

- 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）

数独部分空格内已填入了数字，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：[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]] **解释：**输入的数独如上图所示，唯一有效的解决方案如下所示：

提示：

- `board.length == 9`

- `board[i].length == 9`

- `board[i][j]` 是一位数字或者 `'.'`

- 题目数据 **保证** 输入数独仅有一个解

难度： Hard

题解代码 #

/**
 * @param {character[][]} board
 * @return {void} Do not return anything, modify board in-place instead.
 */

const solveSudoku = board => {
    for (let i = 0; i < 9; i++ ) {
        for (let j = 0; j < 9; j++) {
            if (board[i][j] !== '.') continue
            // 放一个数字试试
            for (let k = 1; k <=9; k++) {
                // 能放
                if (isValid(board, i, j, k.toString())) {
                    board[i][j] = k.toString()
                    if (solveSudoku(board)) return true
                    board[i][j] = '.'
                }
            }
            return false
        }
    }
    return true
}


function isValid (board, row, col, k) {
    const x = Math.floor(row / 3) * 3
    const y = Math.floor(col / 3) * 3
    for (let i = 0; i < 9; i++) {
        if (board[row][i] === k || board[i][col] === k) return false
    }
    for (let i = 0; i < 3; i++) {
        for (let j = 0; j < 3; j++) {
            if (board[x + i][y + j] === k) return false
        }
    }
    return true
}