Ich versuche, die gesamte Zeitkomplexität dieser Funktion mit Big-Oh-Notation zu finden, Die Funktion checkElements() wird rekursiv aufgerufen, die innerhalb der Percolates() liegt vielZeitkomplexität der rekursiven Funktion finden
geschätztpublic static boolean percolates(boolean[][] open) {
size = open.length;
uf = new WeightedQuickUnionUF((size * size) + 2);
for (int i = 0; i < open.length; i++) {//connect all top row elements to virtual top node.
uf.union(0, i);
}
for (int j = 0; j < open.length; j++) {//connect all bottom row elements to bottom virtual node
uf.union((size * size) + 1, (size * size) - j);
}
int row = 0; // current row of grid
int column = 0;// current column of grid
int ufid = 1; // current id of union find array
checkElements(column, row, open, ufid);
boolean systemPerculates = uf.connected(0, (size * size) + 1);
System.out.println("Does the system percoloates :" + systemPerculates);
return systemPerculates;
}
//search elements in the grid
public static void checkElements(int column, int row, boolean open[][], int ufid) {
if (open[row][column]) {
if (column - 1 >= 0 && open[row][column - 1]) { //check adjacent left
uf.union(ufid, ufid - 1);
}
if (column + 1 < size && open[row][column + 1]) {//check adjacent right
uf.union(ufid, ufid + 1);
}
if (row - 1 >= 0 && open[row - 1][column]) {//check adjacent top
uf.union(ufid, ufid - size);
}
if (row + 1 < size && open[row + 1][column]) {//check adjacent bottom
uf.union(ufid, ufid + size);
}
}
if (column + 1 < size) { //go to next column
ufid++;
column++;
checkElements(column, row, open, ufid);
} else if (column + 1 == size && row + 1 < open.length) { //go to next row
ufid++;
row++;
column = 0;
checkElements(column, row, open, ufid);
} else {
return;
}
}
Wo ist die Rekursion? –
die Methode checkElements() –