在早期编程阶段,我一直在修改一个练习。
以下数独解算器用于在有1个解时打印结果,如果有更多解,则仅打印解的数量(因此在这种情况下不显示解)。问题是:如果我从solveIt()打印(),我会将输入数独作为输出。如果我从solve()中打印(),我总是至少得到一个解。
有没有一种方法可以存储已求解的网格,以便我可以从solveIt()调用它,或者有一种方法可以只打印()解决方案的数量,而不会在只有一个解决方案的情况下失去打印解决方案的能力?
完整代码如下
class Sudoku {
int SIZE = 9; // size of the grid
int DMAX = 9; // maximal digit to be filled in
int BOXSIZE = 3; // size of the boxes (subgrids that should contain all digits)
int[][] grid; // the puzzle grid; 0 represents empty
int rempty = 0;
int cempty = 0;
// a challenge-sudoku from the web
int[][] somesudoku = new int[][] {
{ 0, 6, 0, 0, 0, 1, 0, 9, 4 }, //original; one solution
//{ 0, 0, 0, 0, 0, 1, 0, 9, 4 }, //to get more solutions
{ 3, 0, 0, 0, 0, 7, 1, 0, 0 },
{ 0, 0, 0, 0, 9, 0, 0, 0, 0 },
{ 7, 0, 6, 5, 0, 0, 2, 0, 9 },
{ 0, 3, 0, 0, 2, 0, 0, 6, 0 },
{ 9, 0, 2, 0, 0, 6, 3, 0, 1 },
{ 0, 0, 0, 0, 5, 0, 0, 0, 0 },
{ 0, 0, 7, 3, 0, 0, 0, 0, 2 },
{ 4, 1, 0, 7, 0, 0, 0, 8, 0 },
};
int solutionnr = 0; //solution counter
// ----------------- conflict calculation --------------------
// is there a conflict when we fill in d at position r,c?
boolean givesConflict(int r, int c, int n) {
if (rowConflict(r, n) == true || colConflict(c, n) == true || boxConflict(r, c, n) == true) {
return true;
}
return false;
}
boolean rowConflict(int r, int n) {
for (int i = 0; i < 9; i++) {
if (grid[r][i] == n) {
return true;
}
}
return false;
}
boolean colConflict(int c, int n) {
for (int i = 0; i < 9; i++) {
if (grid[i][c] == n) {
return true;
}
}
return false;
}
boolean boxConflict(int rr, int cc, int n) {
int r = (rr / 3) * 3;
int c = (cc / 3) * 3;
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
if (grid[r + i][c + j] == n) {
return true;
}
}
}
return false;
}
// --------- solving ----------
// finds the next empty square (in "reading order")
// writes the coordinates in rempty and cempty
// returns false if there is no empty square in the current grid
int[] findEmptySquare() {
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
if (grid[i][j] == 0) {
int [] coordinateOfEmptySquare = {i, j};
rempty = i;
cempty = j;
return coordinateOfEmptySquare;
}
}
}
solutionnr++;
return null;
}
// prints all solutions that are extensions of current grid
// leaves grid in original state
void solve() {
// if there are no empty squares left we print the current solution
if (findEmptySquare() == null) {
return;
}
int r = rempty;
int c = cempty;
for (int i = 1; i <= 9; i++) {
if (!givesConflict(r, c, i)) {
grid[r][c] = i;
solve();
grid[r][c] = 0;
}
}
//fill in
}
// ------------------------- misc -------------------------
// print the grid, 0s are printed as spaces
void print() {
for (int i = 0; i < 10; i++) {
if (i == 0 || i == 9) {
System.out.println("+-----------------+");
}
if (i == 3 || i == 6) {
System.out.println("-------------------");
}
if (i == 9) {
break;
}
System.out.print("|");
for (int j = 0; j < 3; j++) {
if (j == 1 || j == 2) {
System.out.print(" ");
}
if (grid[i][j] == 0) {
System.out.print(" ");
} else { System.out.print(grid[i][j]);
}
}
System.out.print("|");
for (int j = 3; j < 6; j++) {
if (j == 4 || j == 5) {
System.out.print(" ");
}
if (grid[i][j] == 0) {
System.out.print(" ");
} else { System.out.print(grid[i][j]);
}
}
System.out.print("|");
for (int j = 6; j < 9; j++) {
if (j == 7 || j == 8) {
System.out.print(" ");
}
if (grid[i][j] == 0) {
System.out.print(" ");
} else { System.out.print(grid[i][j]);
}
}
System.out.print("|");
System.out.println();
}
//fill in
}
// --------------- where it all starts --------------------
void solveIt() {
grid = somesudoku;
solve();
print();
if (solutionnr > 1) {
System.out.println(solutionnr);
}
//fill in
}
public static void main(String[] args) {
new Sudoku().solveIt();
}