我正在使用数独解算器,但无法正确返回/结束解算器功能。 moveOn()
函数中的 show()
被调用,它会很好地显示已完成的数独,但solve 返回 false。我试图在问题解决时让解决返回 true,在无法解决时返回 null,但不知道如何实现这一点。
L
是棋盘的长度(9 x 9 数独的 L = 9)
getSquare(r,c)
返回表示数独板的二维数组中的值
不同的检查函数检查某个值是否适合特定位置。它们不是问题。
show()
函数在控制台中打印出数组,因此它看起来像一个正确的数独板。
我还有一个 isSolved()
函数来检查二维数组,如果它是有效的已解决数独,则返回 true,否则返回 false。我也尝试将其实现到 solve()
方法中,希望使用它返回 true,但没有成功
//This method's only purpose it to call the findNum function on the next location in the sudoku
public void moveOn(int row, int column) {
//if the previous location was not the last in the row move to ne next cell in said row.
//if it was the last location in the row, move to the first column of the next row
if (column + 1 != L) {solve(row, column + 1);}
else if (row + 1 != L) {solve(row + 1, 0);}
else {show();}
}
//This method finds a valid number for a specific location on the sudoku grid\
public boolean solve(int row, int column) {
if (row >= L) {return true;}
//pass over any numbers that are not empty
if (getSquare(row, column) != 0) {moveOn(row, column);}
else {
//attempt to find a valid number for the location
for (int n = 1; n <= L; n++) {
if (checkRow(row, n) && checkCol(column, n) && checkSquare(row, column, n)) {
// If a number is allowed at a specific location set that location to the number
setSquare(row, column, n);
//Begin checking for a solution based on previous numbers changed
moveOn(row, column);
}
}
//If no number is allowed in this space backtrack to the last successful number
//changed and reset all locations that have been changed recursively
setSquare(row, column, 0);
}
//If the puzzle is unsolveable
return false;
}
非常感谢任何可以帮助阐明这一情况的人。 如果需要更多我的代码/信息,我很乐意提供
示例输入文件:http://pastebin.com/6mSKT3ES
编辑:删除完整代码
最佳答案
solve
函数中只有一个 return
语句,即
return false;
由于这是函数中的最后一个语句,并且是无条件执行的,因此 solve
将始终返回 false
,除非引发异常。
要获得实际告诉您是否找到解决方案的返回值,您需要使返回值取决于条件。另外,一旦你找到了解决方案,对于适定谜题,继续搜索就没有意义了。
所以你应该在搜索循环中添加一个条件return true;
。为此,您需要知道何时找到解决方案。您将递归包装在对 moveOn
的中间调用中,因此最简单的更改就是向 moveOn
添加返回值:
public boolean moveOn(int row, int column) {
//if the previous location was not the last in the row move to ne next cell in said row.
//if it was the last location in the row, move to the first column of the next row
if (column + 1 != L) {return solve(row, column + 1);}
else if (row + 1 != L) {return solve(row + 1, 0);}
else {show(); return true;} // reached end of grid, solved
}
并在“解决”中使用它:
public boolean solve(int row, int column) {
//pass over any numbers that are not empty
if (getSquare(row, column) != 0) {return moveOn(row, column);}
else {
//attempt to find a valid number for the location
for (int n = 1; n <= L; n++) {
if (checkRow(row, n) && checkCol(column, n) && checkSquare(row, column, n)) {
// If a number is allowed at a specific location set that location to the number
setSquare(row, column, n);
//Begin checking for a solution based on previous numbers changed
if (moveOn(row, column)) {
return true; // solved, yay!
}
}
}
//If no number is allowed in this space backtrack to the last successful number
//changed and reset all locations that have been changed recursively
setSquare(row, column, 0);
}
//If the puzzle is unsolveable
return false;
}
关于Java递归数独求解器,递归函数不返回正确值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12964564/