[Day43]-[回溯]- 解数独
作者:方勇(gopher)
- 2022 年 5 月 19 日
本文字数:1202 字
阅读完需:约 4 分钟
37. 解数独
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:
复制代码
题解:
func TestSolveSudoku(t *testing.T) {
var m = 9 // 9x9 的格子
var board [][]byte // 数独
var isValid func(row, col int, target byte) bool // 是否合法
var traceback func(row, col int) bool // 回溯
isValid = func(row, col int, target byte) bool {
for i := 0; i < m; i++ {
if board[row][i] == target { // 同一行不能重复
return false
}
if board[i][col] == target { // 同一列不能重复
return false
}
if board[row/3*3+i/3][col/3*3+i%3] == target { //3x3区域不能重复
return false
}
}
return true
}
// 初始化棋盘
func() {
board = make([][]byte, m)
for i := 0; i < m; i++ {
board[i] = make([]byte, m)
}
for i := 0; i < m; i++ {
for j := 0; j < m; j++ {
board[i][j] = '.'
}
}
}()
traceback = func(row, col int) bool {
if col == m {
return traceback(row+1, 0)
}
if row == m {
return true
}
if board[row][col] != '.' {
return traceback(row, col+1)
}
for i := '1'; i <= '9'; i++ {
if !isValid(row, col, byte(i)) {
continue
}
board[row][col] = byte(i)
if traceback(row, col+1) {
return true
}
board[row][col] = '.'
}
return false
}
traceback(0, 0)
for i := 0; i < m; i++ {
fmt.Println()
for j := 0; j < m; j++ {
fmt.Print(string(board[i][j]))
}
}
}
复制代码
划线
评论
复制
发布于: 刚刚阅读数: 2
方勇(gopher)
关注
Dead or Alive. 生存战斗是知识的源泉! 2018.11.08 加入
我是一名SRE哨兵,目前是好大夫基础架构部高级工程师。专注于 SRE,微服务、中间件的稳定性和可用性建设,整体负责好大夫服务治理云平台的设计和搭建!
评论