LeetCode - 37 解数独

前言

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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

[图片上传失败…(image-122cc7-1649473296178)]

**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：

[图片上传失败…(image-bfd31a-1649473296178)]

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
    }
}

主要思想：遍历整个矩阵，尝试用所有可能的情况填满空白，并检查有效性。
时间复杂度： O(n^4)
空间复杂度： O(1)

该算法题解的仓库：LeetCode-Swift

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