Go语言实现的最简单数独解法

package main
import (
    "fmt"
)
type node []int
var sudokuMay [9][9]node
var Sudoku = [9][9]int{
    {0, 0, 0, 0, 0, 0, 8, 0, 0},
    {0, 8, 2, 4, 0, 0, 0, 0, 0},
    {1, 9, 0, 0, 6, 3, 0, 0, 0},
    {0, 5, 0, 0, 8, 0, 7, 0, 0},
    {6, 7, 8, 2, 0, 9, 1, 4, 3},
    {0, 0, 3, 0, 4, 0, 0, 8, 0},
    {0, 0, 0, 6, 2, 0, 0, 9, 4},
    {0, 0, 0, 0, 0, 5, 6, 1, 0},
    {0, 0, 0, 6, 0, 0, 0, 0, 0}}
func main() {
    n := inited(Sudoku)
    SudokuSure, _ := sure(sudokuMay)
    for n > 0 {
        n = Subinit(SudokuSure)
        // Output(sudokuMay)
        // fmt.Println(n)
        SudokuSure, _ = sure(sudokuMay)
    }
    Output(sudokuMay)
    fmt.Println(isEnable(sudokuMay))
    // test()
}
func isEnable(tn [9][9]node) bool {
    for i := 0; i < 9; i++ {
        for j := 0; j < 9; j++ {
            if len(tn[i][j]) == 0 {
                return false
            }
        }
    }
    return true
}
func sure(may [9][9]node) (sure [9][9]int, n int) {
    n = 0
    for i := 0; i < 9; i++ {
        for j := 0; j < 9; j++ {
            if len(may[i][j]) == 1 {
                sure[i][j] = may[i][j][0]
                n++
            } else {
                sure[i][j] = 0
            }
        }
    }
    return
}
func test() {
    i, j := 1, 3
    fmt.Println(Sudoku[i][j])
    for k := ((i / 3) * 3); k < ((i/3)*3)+3; k++ {
        for l := ((j / 3) * 3); l < ((j/3)*3)+3; l++ {
            fmt.Print(Sudoku[k][l])
        }
        fmt.Println(" ")
    }
}
func inited(Sud [9][9]int) (changeCount int) {
    tmp := 0
    changeCount = 0
    for i := 0; i < 9; i++ {
        for j := 0; j < 9; j++ {
            if Sud[i][j] != 0 {
                sudokuMay[i][j] = append(sudokuMay[i][j], Sud[i][j])
            } else {
                for k := 0; k < 9; k++ {
                    sudokuMay[i][j] = append(sudokuMay[i][j], k+1)
                }
                sudokuMay[i][j], tmp = excludeMay(i, j, sudokuMay[i][j], Sud)
                changeCount += tmp
            }
        }
    }
    return
}
func Subinit(Sud [9][9]int) (changeCount int) {
    tmp := 0
    changeCount = 0
    for i := 0; i < 9; i++ {
        for j := 0; j < 9; j++ {
            if Sud[i][j] != 0 {
                sudokuMay[i][j][0] = Sud[i][j]
            } else {
                sudokuMay[i][j], tmp = excludeMay(i, j, sudokuMay[i][j], Sud)
                changeCount += tmp
            }
        }
    }
    return
}
func excludeMay(ti, tj int, t node, S [9][9]int) (rmay node, changeCount int) {
    changeCount = 0
    var tmpChangeCount int
    for i := 0; i < 9; i++ {
        if S[i][tj] != 0 {
            t, tmpChangeCount = exclude(t, S[i][tj])
            changeCount += tmpChangeCount
        }
        if S[ti][i] != 0 {
            t, tmpChangeCount = exclude(t, S[ti][i])
            changeCount += tmpChangeCount
        }
    }
    for k := ((ti / 3) * 3); k < ((ti/3)*3)+3; k++ {
        for l := ((tj / 3) * 3); l < ((tj/3)*3)+3; l++ {
            if S[k][l] != 0 {
                t, tmpChangeCount = exclude(t, S[k][l])
                changeCount += tmpChangeCount
            }
        }
    }
    rmay = t
    return
}
func excludeFirstOne(smay node, n int) (rmay node, changeCount int) {
    changeCount = 0
    rmay = smay
    for i := 0; i < len(smay); i++ {
        if smay[i] == n {
            changeCount++
            rmay = append(smay[:i], smay[i+1:]...)
            return
        }
        if i == len(smay)-1 {
            return
        }
    }
    return
}
func exclude(smay node, n int) (tmp node, changeCount int) {
    var nc int
    changeCount = 0
    tmp, nc = excludeFirstOne(smay, n)
    for nc > 0 {
        tmp, nc = excludeFirstOne(tmp, n)
        changeCount++
    }
    return
}
func Output(sudoku [9][9]node) {
    for i := 0; i < 9; i++ {
        for j := 0; j < 9; j++ {
            fmt.Print(sudokuMay[i][j])
        }
        fmt.Println("")
    }
}