问题
Program Description:
Write a program to print 21 rows of X's in the shape of a large X as illustrated below. Be sure so the two rows intersect at the "11" row.
Here is what I want as an output:
Here is what I have so far.
public class Program168h {
public static void main (String [] args) {
String d= "X";
for (int a = 1; a < 23; a++) {
for (int b = a; b >= 1; b--) {
System.out.print(" ");
}
System.out.print(d);
for (int x = a; x < 22; x++) {
System.out.print(" ");
}
System.out.print(d);
System.out.println();
}
}
}
This only produces the first half of the X, I do not know how to produce the lower half.
回答1:
Try this one:
int xSize = 21;
int ySize = 21;
String sign = "X";
for (int i = 0; i < xSize; ++i) {
for (int j = 0; j < ySize; ++j) {
if (i == j) {
System.out.print(sign);
} else if (i == ySize - j - 1) {
System.out.print(sign);
} else {
System.out.print(" ");
}
}
System.out.println();
}
explanation: The first operate on Xaxis coordinates, second for operates on Yaxis. Our task is to cover diagonal. Covering first diagonal is where coordinateX == coordinateY. In code is if(i==j). These are points (1,1), (2,2)...... Second diagonal are points where (x,y)= (20,1),(19,2),(18,3) .... This situation covers second if(i == ySize - j - 1) .
回答2:
You can try that:
public class ProductX {
public static void main(String[] args) {
for (int i = 0; i <= 10; i++) {
for (int j = 0; j < 10; j++) {
System.out.print(" ");
if (i == j) {
System.out.print("X");
}
if(j == 9-i){
System.out.print("X");
}
}
System.out.println();}
}
}
回答3:
Though the above solutions work perfectly, I tried to experiment by not using nested for and the soultion is as below. This will have higher performance than using nested for which has a complexity of O(n2) when compared to O(n) in this.
public void testXFormation() {
final int countOfLines = 21;
int countOfSpaceBefore = 0;
int countOfSpacesAfter = countOfLines -2 ;// 2 characters
boolean halfReached = false;
for (int index = 0; index < countOfLines; index++) {
printSpaces(countOfSpaceBefore); // print required no. of spaces
System.out.print("x"); // print first x
printSpaces(countOfSpacesAfter); // print required no. of spaces after x
if (index != (countOfLines / 2))// Avoid printing double, in the middle
System.out.print("x");
System.out.println(""); // move to next line
/* Once you reach half matrix we need to reverse the logic */
if (index >= (countOfLines - 1) / 2) {
halfReached = true;
}
/* Reversing the logic for the spaces to be printed */
if (halfReached) {
countOfSpaceBefore--;
countOfSpacesAfter += 2;
} else {
countOfSpaceBefore++;
countOfSpacesAfter -= 2;
}
}
}
private void printSpaces(int count) {
for (int i = 0; i < count; i++)
System.out.print(" ");
}
来源:https://stackoverflow.com/questions/33173715/java-nested-loops