为什么数独求解器的这种递归方法不断旋转？
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在学校的作业中，我们应该做一个soduko求解器。我有一个递归方法，应该可以帮助我解决soduko难题。它是这样的：
public void setNumber() {
if (getNext() == null) {
for (int i = 1; i < 10; i++) {
if (acceptValue(i)) {
value = i;
}
}
board.stopRec();
return;
} else {
if(predefined()) { // if the square allready has a number
getNext().setNumber();
} else { // find value for undefined square
for (int i = 1; i < 10; i++) {
if (acceptValue(i)) {
value = i // Does the Row class take notice of this?
getNext().setNumber();
}
}
/* no value was assigned */
value = -1;
}
}
}
class Board {
[...]
public void initGrid(int nSquaresBoxRowCol, int nRowsBox, int nColsBox, int[][] values) {
[...]
/* Create the rows, columns and squares */
Square prev = null;
for (int row = 1; row <= nRowsBoard; row++) {
//Row r = new Row(row, squares[row-1]);
Row r = new Row(row);
rows[row-1] = r;
for (int col = 1; col <= nColsBoard; col++) {
if (row == 1) {
Column c = new Column(col);
}
Square current = new Square(row, col, Math.ceil((float)row / (float)nRowsBox), Math.ceil((float)col / (float)nColsBox), values[row-1][col-1], r, this);
if (!((row-1) == 0 && (col-1) == 0)) {
prev.nextSquare = current;
}
prev = current;
squares[row-1][col-1] = current;
r.addSquare(current);
/* Fill the boxes with squares */
boxes[(int)(Math.ceil((float)row / (float)nRowsBox)) - 1][(int)(Math.ceil((float)col / (float)nColsBox)) - 1].addSquare(squares[row-1][col-1]);
nSquares++;
}
} // END for (int row ...
class Row {
int id;
Square[] squares;
ArrayList<Square> squareList;
Row(int id) {
this.id = id; // not currently used for anything
squareList = new ArrayList<Square>();
}
boolean checkValue(int val) {
System.out.println(\"Checking values for row \" + id);
Iterator<Square> iter = squareList.iterator();
while(iter.hasNext()) {
System.out.println(\"Value (boolean checkValue() in Row): \" + iter.next().value);
if (iter.next().value == val) {
System.out.println(\"returned false\");
return false;
}
}
return true;
}
public void addSquare(Square s) {
squareList.add(s);
System.out.println(\"class Row, method addSquare: Square with value \" + s.value + \" added to row.\");
}
没有找到相关结果
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1 个回复
室邢
每次都会正确选择，那么您会想在之后返回 就像是for(i=1;i<10;i++){ if(acceptValue(i)){ value = i; getNext().setNumber(); return; } }返回时选择2作为值，它将尝试3、4、5 ...，依此类推。