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[1]）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。
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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
**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：
约束条件：
board.length == 9board[i].length == 9board[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
代码语言：javascript复制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[2]
点击前往 LeetCode[3] 练习
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特别感谢 Swift社区 编辑部的每一位编辑，感谢大家的辛苦付出，为 Swift社区 提供优质内容，为 Swift 语言的发展贡献自己的力量，排名不分先后：张安宇@微软[4]、戴铭@快手[5]、展菲@ESP[6]、倪瑶@Trip.com[7]、杜鑫瑶@新浪[8]、韦弦@Gwell[9]、张浩@讯飞[10]、张星宇@ByteDance[11]、郭英东@便利蜂[12]
参考资料
[1]
@故胤道长: https://m.weibo.cn/u/1827884772
[2]
LeetCode-Swift: https://github.com/soapyigu/LeetCode-Swift
[3]
LeetCode: https://leetcode.com/problems/sudoku-solver/
[4]
张安宇: https://blog.csdn.net/mobanchengshuang
[5]
戴铭: https://ming1016.github.io
[6]
展菲: https://github.com/fanbaoying
[7]
倪瑶: https://github.com/niyaoyao
[8]
杜鑫瑶: https://weibo.com/u/3878455011
[9]
韦弦: https://www.jianshu.com/u/855d6ea2b3d1
[10]
张浩: https://github.com/zhanghao19920218
[11]
张星宇: https://github.com/bestswifter
[12]
郭英东: https://github.com/EmingK
- EOF -
