[英]Print a rectangle with diagonals using a 2d array in Java
我一直在尝试修改打印到控制台的普通矩形2d数组,以显示其对角线以及其他字符。 例如,我当前具有2d数组的矩形的代码是:
import java.util.Scanner;
class RecArray {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print("Height: ");
int height = scanner.nextInt();
System.out.print("Width: ");
int width = scanner.nextInt();
char[][] square = new char[height][width];
String line;
// fill the array
for (int i = 0; i < height; i++) {
for (int j = 0; j < width; j++) {
square[i][j] = 'o';
}
}
// print the array
for (int i = 0; i < height; i++) {
line = "";
for (int j = 0; j < width; j++) {
line += square[i][j];
}
System.out.println(line);
}
}
}
它返回:
Height: 10
Width: 10
oooooooooo
oooooooooo
oooooooooo
oooooooooo
oooooooooo
oooooooooo
oooooooooo
oooooooooo
oooooooooo
oooooooooo
我希望对角线代码返回:
Height: 5
Width: 7
xooooox
oxoooxo
ooxxxoo
oxoooxo
xooooox
我当前的代码是:
import java.util.Scanner;
class RecArrayDiag {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print("Height: ");
int height = scanner.nextInt();
System.out.print("Width: ");
int width = scanner.nextInt();
char[][] square = new char[height][width];
boolean bool1 = true;
boolean bool2 = true;
boolean bool3 = true;
boolean bool4 = true;
String line;
int x = 0;
for (int i = 0; i < height; i++) {
for (int j = width-1; j >= 0; j--) {
if (i % 2 == 0 ? ((i == height/2)) : ((i == height-1/2))) {
bool1 = false;
}
if (j % 2 == 0 ? ((j == width/2)) : ((j == width-1/2))) {
bool2 = false;
}
if ((((i == j) && bool1 && bool2) || (i == (height - (j+1))) || (j == (width - (i+1))) || ((j == width-1) && bool3) || ((i == height-1) && bool4) || (j == width-1) && (i == height-1))) {
square[i][j] = 'x';
//x++;
} else {
square[i][j] = 'o';
}
if ((j == width-1)) {
bool3 = false;
}
if ((i == height-1)) {
bool4 = false;
}
}
x++;
}
// print the array
for (int i = 0; i < height; i++) {
line = "";
for (int j = 0; j < width; j++) {
line += square[i][j];
}
System.out.println(line);
}
}
}
并返回:
Height: 5
Width: 7
xoooxox
oxoxoxo
ooxoxoo
oxoxooo
xoxooox
请帮助我解决这个问题,并在此先感谢。
这是一种实现方法,可重用的方法将应用于矩形的各种操作分开。
public static void printRectangleWithDiagonals(int width, int height) {
char[][] rectangle = new char[height][width];
fill(rectangle, 'o');
drawDiagonals(rectangle, 'x');
print(rectangle);
}
private static void fill(char[][] rectangle, char ch) {
for (char[] line : rectangle)
for (int i = 0; i < line.length; i++)
line[i] = ch;
}
private static void drawDiagonals(char[][] rectangle, char ch) {
int bottom = rectangle.length - 1, right = rectangle[0].length - 1;
if (right > bottom) {
for (int x = 0; x <= right; x++) {
int y = (x * bottom + right / 2) / right;
rectangle[y][x] = ch;
rectangle[bottom - y][x] = ch;
}
} else {
for (int y = 0; y <= bottom; y++) {
int x = (y * right + bottom / 2) / bottom;
rectangle[y][x] = ch;
rectangle[y][right - x] = ch;
}
}
}
private static void print(char[][] rectangle) {
for (char[] line : rectangle)
System.out.println(line);
}
测试
printRectangleWithDiagonals(7, 7);
System.out.println();
printRectangleWithDiagonals(10, 4);
System.out.println();
printRectangleWithDiagonals(5, 9);
产量
xooooox
oxoooxo
ooxoxoo
oooxooo
ooxoxoo
oxoooxo
xooooox
xxooooooxx
ooxxxxxxoo
ooxxxxxxoo
xxooooooxx
xooox
oxoxo
oxoxo
ooxoo
ooxoo
oxoxo
oxoxo
xooox
xooox
据我了解,您想展示某种十字架。 您要处理矩阵不是正方形的情况。
这意味着您可以直接从所有角度到达中心点,如果一个轴到达数组的中间,则首先停止计数器并继续第二个参数。
像这样的东西(只是伪代码):
//create square with "o" everywhere then overwrite
int i = 0;
int j = 0;
while(i < height/2 || j < width/2){
//go from all corners towards the middle
if (i == j){
square[i][j] = "x";
square[i][width - j+1] = "x";
square[height - i+1][j] = "x";
square[height - i+1][width - j+1] = "x";
} else if (i < height/2) { //i is in middle of array
square[i][j] = "x";
square[i][width - j+1] = "x";
} else { //j is is in middle of array
square[i][j] = "x";
square[height - i+1][j] = "x";
}
//as long i and j did not reach the center add 1
if (i < width/2) { i++ }
if (j < height/2) { j++ }
}
希望这个对你有帮助。 通常,我建议将您的问题分解为不同的部分。
我可以在您的解决方案中看到逻辑,但请尝试使其保持简单。 查找只要条件为真的规则。 (在这种情况下:只要您不在任何数组中间),然后尝试为不正确的情况找到解决方案。 (例如,如果我到达数组的中间但j没有,会发生什么)
这样,您可以拆分代码,使其更易于阅读/维护。
在大多数情况下,如果您有大量的if else语句,则很有可能将它们重写为较小的部分。
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