LeetCode - #37 解数独

原创

网罗开发 2022-09-01 14:26:30 ©著作权

文章标签 LeetCode 算法 swift 数独数据 文章分类 后端开发

©著作权归作者所有：来自51CTO博客作者网罗开发的原创作品，请联系作者获取转载授权，否则将追究法律责任

前言

我们社区陆续会将顾毅（Netflix 增长黑客，《iOS 面试之道》作者，ACE 职业健身教练。微博:@故胤道长）的 Swift 算法题题解整理为文字版以方便大家学习与阅读。

LeetCode 算法到目前我们已经更新了 36 期，我们会保持更新时间和进度（周一、周三、周五早上 9:00 发布），每期的内容不多，我们希望大家可以在上班路上阅读，长久积累会有很大提升。

不积跬步，无以至千里；不积小流，无以成江海，Swift社区伴你前行。如果大家有建议和意见欢迎在文末留言，我们会尽力满足大家的需求。

难度水平：困难

1. 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。

2. 示例

LeetCode - #37 解数独_LeetCode

**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 = [["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：

LeetCode - #37 解数独_LeetCode_02

约束条件：

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

3. 答案

class SudokuSolver{
    func solveSudoku(_ board: inout [[Character]]) {
        guard board.count != 0 || board[0].count != 0 else {
            return
        }
        helper(&board)
    }

    private func helper(_ board: inout [[Character]]) -> Bool {
        let m = board.count, n = board[0].count

        for i in 0..<m {
            for j in 0..<n {
                if board[i][j] == "." {
                    for num in 1...9 {
                        if isValid(board, i, j, Character(String(num))) {
                            board[i][j] = Character(String(num))

                            if helper(&board) {
                                return true
                            } else {
                                board[i][j] = "."
                            }
                        }
                    }
                    return false
                }
            }
        }

        return true
    }

    private func isValid(_ board: [[Character]], _ i: Int, _ j: Int, _ num: Character) -> Bool {
        let m = board.count, n = board[0].count

        // check row
        for x in 0..<n {
            if board[i][x] == num {
                return false
            }
        }

        // check col
        for y in 0..<m {
            if board[y][j] == num {
                return false
            }
        }

        // check square
        for x in i / 3 * 3..<i / 3 * 3 + 3 {
            for y in j / 3 * 3..<j / 3 * 3 + 3 {
                if board[x][y] == num {
                    return false
                }
            }
        }

        return true