Description
tags: backtrack,hash table difficulty: hard
Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy all of the following rules:
Each of the digits 1-9 must occur exactly once in each row.
Each of the digits 1-9 must occur exactly once in each column.
Each of the the digits 1-9 must occur exactly once in each of the 9 3x3 sub-boxes of the grid.
Empty cells are indicated by the character '.'.
A sudoku puzzle...
...and its solution numbers marked in red.
Note:
The given board contain only digits 1-9 and the character '.'.
You may assume that the given Sudoku puzzle will have a single unique solution.
The given board size is always 9x9.
Solution
package main
import "fmt"
func main() {
board := [][]byte{{'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'}}
solvedBoard := [][]byte{{'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'}}
fmt.Println(solvedBoard)
solveSudoku(board)
fmt.Println(board)
}
func solveSudoku(board [][]byte) {
backtrack(board,0,0)
}
func threeThreeLoc(i,j,r int)(int, int){
if i <= 2 && j <= 2 {
return r / 3, r % 3
} else if i <=2 && j <= 5 {
return r / 3 , r%3 + 3
} else if i<= 2{
return r / 3 , r%3 + 6
}
if i <= 5 && j <=2 {
return r/3+3 , r%3
} else if i <=5 && j <=5 {
return r/3+3 , r%3 + 3
} else if i<=5{
return r/3+3, r%3 +6
}
if i <= 9 && j <=2 {
return r/3+6 , r%3
} else if i <=9 && j <=5 {
return r/3+6 , r%3 + 3
} else {
return r/3+6, r%3 +6
}
}
func backtrack(board [][]byte, i, j int) bool {
m,n := nextBlank(board, i, j)
if m >=9 || n >=9 {
return true
}
for _, value := range []int{1,2,3,4,5,6,7,8,9} {
if value <= 0 || !isValidPlace(board, m, n, value){
continue
}
board[m][n] = byte(value) + '0'
if backtrack(board, m, n){
return true
}
board[m][n] = '.'
}
return false
}
func isValidPlace(board [][] byte, i, j, value int) bool {
for r:=0;r<9;r++{
if int(board[i][r]-'0') == value {
return false
}
if int(board[r][j] - '0') == value {
return false
}
m,n := threeThreeLoc(i, j, r)
if int(board[m][n] - '0') == value {
return false
}
}
return true
}
func nextBlank(board [][]byte, i, j int) (int, int){
for loc := i * 9 + j; loc < 81; loc++ {
m,n := loc / 9, loc % 9
if board[m][n] == '.' {
return m,n
}
}
return 10,10
}
Note:
每一個點的位置都有1-9種可能性，但是不是每一個都滿足需求，因此通過回溯建立了最終的解。這個解法沒有用到Hash Table，在判斷是否是有效值時浪費了時間，並不是一種最優的方案。