I need to print this:
I need to print this square using a for loop in java.
This is what I have so far:
public static void main(String [] args)
{
for(int i = 0; i < 5; i++){
System.out.print("O ");
}
System.out.println();
for(int i = 0; i < 4; i++){
System.out.print("O ");
}
}
The easiest solution would be to nest two loops, first print O then use a second loop to print .. You know that each line should have a decreasing number of O (and increasing number of .s). In fact, you have 5 - i and i of each per line (where i is row number). And you can use the outer loop to determine how many of each should be drawn with those formulae. Like,
int size = 5;
for (int i = 0; i < size; i++) {
for (int j = 0; j < size - i; j++) {
System.out.print("O ");
}
for (int j = 0; j < i; j++) {
System.out.print(". ");
}
System.out.println();
}
Which outputs (as requested)
O O O O O
O O O O .
O O O . .
O O . . .
O . . . .
Another option would be to create a StringBuilder to represent the initial conditions, print and then mutate the StringBuilder. This uses additional storage, but eliminates the need for nested loops.
int size = 5;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < size; i++) {
sb.append("O ");
}
for (int i = 0; i < size; i++) {
System.out.println(sb);
sb.setCharAt(sb.length() - (i * 2) - 2, '.');
}
And, you could also make that with an array of boolean(s) (e.g. a bitmask). Convert false to O and true to ., and set the last element offset by the index to true on each iteration. Like,
int size = 5;
boolean[] arr = new boolean[size];
for (int i = 0; i < size; i++) {
for (int j = 0; j < size; j++) {
System.out.print(arr[j] ? ". " : "O ");
}
System.out.println();
arr[size - i - 1] = true;
}
This is not a 'design pattern' problem. You just need to use the logic with loops. Here's the logic.
In a square of n x n cells, the ith row contains (n-i) ovals and i dots, where 0 <= i < n.
Refer to the other answer to see a working snippet.
