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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。

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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:

  1. 数字​​1-9​​ 在每一行只能出现一次。
  2. 数字​​1-9​​ 在每一列只能出现一次。
  3. 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_算法

**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_swift_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
  • 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

点击前往 LeetCode 练习

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