在 Java 中的模式一文中，在學(xué)習(xí) Java 中的任何程式語(yǔ)言並深入研究高階概念之前，了解循環(huán)的工作原理非常重要。雖然有 3 種類(lèi)型的循環(huán)，分別是 for、while 和 do-while 循環(huán)。每個(gè)循環(huán)根據(jù)程式的具體情況使用，因?yàn)樗鼈儽舜寺杂胁煌榱耸褂酶鞣N循環(huán)需要一些程式邏輯，為此目的，為程式設(shè)計(jì)師提供了模式練習(xí)，因?yàn)樗婕斑壿嫼屯评砟芰Φ氖褂?。例如，它可以在控制臺(tái)螢?zāi)簧狭杏缀螆D形（如三角形、正方形等）、金字塔、各種星形圖案的盒子、數(shù)字和字元樣式。循環(huán)的格式或基本語(yǔ)法可能因一種程式語(yǔ)言而異，但列印這些模式的一般邏輯保持不變。

Java 中的模式範(fàn)例

讓我們透過(guò)一些範(fàn)例來(lái)了解如何用Java繪製圖案

範(fàn)例1：使用數(shù)字列印金字塔的一半。

代碼：

public class Pyramid
{
public static void main(String[] args)
{
int i, j;
?//outermost loop to represent the number of rows which is 5 in this case for(i= 1; i<= 5; i++)
{
?//innermost loop is to print the numbers in the specific rows for (j=1; j<=i; j++)
{
System.out.print(j +" " );
}
System.out.println();
}
}
}

輸出：

在上面的範(fàn)例中，只需要 2 個(gè)基本迴圈即可列印圖案；第一個(gè) for 迴圈用於計(jì)算行數(shù)。在我們的例子中，我們定義了行，即 5 行，否則我們也可以從使用者那裡獲取輸入並將其儲(chǔ)存在變數(shù)中。內(nèi)循環(huán)是列印特定行中的數(shù)字；完成 1 行或「j」循環(huán)結(jié)束後，使用 println() 變更該行。

範(fàn)例2：列印數(shù)字箭頭。

代碼：

public class NumberTriangle
{
public static void main(String[] args)
{
int i, j;
int rows =7;
?//outermost loop to represent the number of rows which is 7 in this case
//for the upper half of arrow
for (i=1; i<= rows; i++)
{
?//innermost loop is to print the numbers in the specific rows
//for the upper half of arrow
for (j=1; j<=i; j++)
{
System.out.print(j + " ");
}
System.out.println();
}
?//outermost loop to represent the number of rows which is 6 in this case
//for the lower half of arrow
for (i=rows-1; i>=1; i--)
{
?//innermost loop is to print the numbers in the specific rows
//for the lower half of arrow
for (j=1; j<=i; j++)
{
System.out.print(j + " ");
}
System.out.println();
}
}
}

輸出：

?

在上面的範(fàn)例中，我們需要將箭頭分成兩半，並為每一半使用 2 個(gè)循環(huán)。行的前半部將是為行設(shè)定的初始值，而下半部的行計(jì)數(shù)比初始值少 1。兩半的內(nèi)循環(huán)用於根據(jù)外循環(huán)迭代每一行。

範(fàn)例3：使用星號(hào)(*)印出完整的金字塔。

代碼：

public class FullPyramid
{
public static void main(String[] args)
{
int i, j, k;
int rows = 5;
//outermost loop to represent the number of rows which is 5 in this case for(i= 1; i<= rows; i++)
{
//innermost loop to represent the spaces in pyramid for (j= 1; j<= rows-i; j++)
{
System.out.print(" ");
}
?//innermost loop to represent the stars (*) in pyramid for (k= 1; k<= 2*i-1; k++)
{
System.out.print("* ");
}
System.out.println();
}
}
}

輸出：

在上面的範(fàn)例中，我們需要做 3 件事，也就是記住第一個(gè) for 迴圈從 1 到 rows 變數(shù)印出金字塔的總行數(shù)。其次，我們首先需要列印金字塔中的空格，然後列印空格後面的圖案（*）。對(duì)於第二個(gè)和第三個(gè)，for迴圈在外循環(huán)「i」內(nèi)使用。

範(fàn)例 4：使用數(shù)字列印半倒金字塔。

代碼：

public class ReversePyramid
{
public static void main(String[] args)
{
int i, j, k;
int rows = 5;
?//outermost loop to represent the number of rows which is 5 in this case for(i= 1; i<= rows; i++)
{
//innermost loop to represent the spaces
for (j= 1; j<= rows-1; j++)
{
System.out.print(" ");
}
?//innermost loop to represent the stars (*) in pyramid for (k= 1; k<= i; k++)
{
System.out.print("* ");
}
System.out.println();
}
}
}

輸出：

簡(jiǎn)單的半金字塔很容易，因?yàn)槲覀冃枰幚頂?shù)字、*或我們正在打印的字符，但對(duì)於反向金字塔，我們需要先打印空格，然後打印模式，在我們的例子中是 (*) 。因此使用了 3 個(gè) for 循環(huán)，其工作原理與完整金字塔的情況類(lèi)似。

範(fàn)例 5：使用字母列印半個(gè)金字塔。

代碼：

public class AlphabetPyramid
{
public static void main(String[] args)
{
int i, j;
?//outermost loop to represent the number of rows which is 5 in this case for(i= 1; i<= 5; i++)
{
int ch = 65;
?//innermost loop to represent the alphabets in a pyramid in particular row for (j= 1; j<= i; j++)
{
System.out.print((char)(ch + i - 1) + " ");
}
System.out.println();
}
}
}

輸出：

金字塔的列印邏輯與上面範(fàn)例中使用的邏輯相同，使用 2 個(gè) for 循環(huán)，一個(gè)用於行數(shù)，其他用於特定行中的字元列印。但主要要注意的是字元資料的處理。例如Java中‘A’的數(shù)值是65，所以所有的數(shù)學(xué)邏輯都是用字母的數(shù)值進(jìn)行的，最後以字元格式列印出來(lái)。

例6：字母的印刷圖案。

代碼：

public class AlphabetPattern
{
public static void main(String[] args)
{
int i, j;
//outermost loop to represent the number of rows which is 5 in this case for(i= 1; i<= 5; i++)
{
int ch = 65;
?//innermost loop to represent the alphabets for (j= 1; j<= i; j++)
{
System.out.print((char)(ch - 1 + j) + " ");
}
System.out.println();
}
}
}

輸出：

上面範(fàn)例中處理字元值和 2 個(gè) for 迴圈所遵循的基本模式與範(fàn)例 5 類(lèi)似，唯一的差異是用於列印所需模式的簡(jiǎn)單邏輯。

範(fàn)例 7：使用星號(hào) (*) 列印正方形。

代碼：

public class SquarePattern
{
public static void main(String[] args)
{
int i, j;
?//outermost loop to represent the number of rows which is 5 in this case for(i= 1; i<= 5; i++)
{
int ch = 65;
//innermost loop to represent the stars (*) for (j= 1; j<= 5; j++)
{
System.out.print(" * " + " ");
}
System.out.println();
}
}
}

輸出：

For printing of square, we need length and width, i.e. both sides of the square should be the same, which is 5 in our case. So the first ? ?loop is used for the length or number of rows in the square, and the inner ? ?loop is used for the width of the square, i.e. 5 stars in a single row.

Example 8: Printing rectangle using stars (*).

Code:

public class RectanglePattern
{
public static void main(String[] args)
{
int i, j;
?//outermost loop to represent the number of rows which is 5 in this case for(i= 1; i<= 5; i++)
{
int ch = 65;
?//innermost loop to represent columns the stars (*) for (j= 1; j<= 9; j++)
{
System.out.print(" * " + " " );
}
System.out.println();
}
}
}

Output:

The basic logic of printing the rectangle of (*) is the same as printing of squares, the only difference between is the different length and width of the rectangle. Here ‘i’ loop is for the length of the rectangle, and the inner ‘j’ loop is for the width of the loop. Our program is taken as a constant value; we can also ask the user and store them in separate variables.

Example 9: Printing a Diamond using stars.

Printing a diamond in Java is a very simple process. It involves printing 2 pyramids, 1 in the upward direction and another in an inverted direction. Basically, we need to use the loops to do the coding to print two separate pyramids.

Code:

public class Diamond
{
public static void main(String[] args)
{
int i, j, k;
int rows = 5;
?//outermost loop to represent the number of rows which is 5 in this case.
// Creating upper pyramid
for(i= 1; i<= rows; i++)
{
//innermost loop to represent the spaces in upper pyramid for (j= 1; j<= rows-i; j++)
{
System.out.print(" ");
}
?//innermost loop to represent the stars (*) in upper pyramid for (k= 1; k<= 2*i-1; k++)
{
System.out.print("* ");
}
System.out.println();
}
?//outermost loop for the rows in the inverted pyramid for (i = rows-1; i>0; i--)
{
?//innermost loop for the space present in the inverted pyramid for (j=1; j<= rows - i; j++)
{
System.out.print(" ");
}
?//innermost loop inside the outer loop to print the ( * ) pattern in inverted pyramid for (k = 1; k<= 2*i-1; k++)
{
System.out.print("* ");
}
System.out.println();
}
}
}

In the above example, almost the same logic is applied to create both pyramids, one in an upward direction and another in an inverted direction. Thus, the first ?loop is for the number of lines or rows in the pattern, and the second is for spaces and the stars (*) pattern in the pattern.

Output:

Example 10: Printing binary numbers in a stair format.

Code:

public class BinaryStair
{
public static void main(String[] args)
{
int i, j;
//outer loop for the total rows which is 5 in this case for (i = 1; i <= 5; i++)
{
?//inner loop for the pattern of 0 and 1 in each row for (j = 1; j<= i ; j++)
{
if (j % 2 ==0)
{
System.out.print(0);
}
else
{
System.out.print(1);
}
}
System.out.println();
}
}
}

Output:

In the above example, in order to print binary pattern, outer ?for ?loop ‘i’ is used for a total number of rows, and the inner ?for ?loop ‘j’ is used to iterate till the outer loop ‘i’ because for the 1st row, we need 1 value, for the 2nd row we need 2 values, and so on. ?If? and else ?statements are used in order to print the alternate value of 0 and 1. Suppose for the first time i=1, j=1 and 1%2 != 0, then 1 is printed, and execution will move out of the inner loop.

Example 11: Program to print repeating alphabet patterns.

Code:

public class AlphabetReverseOrder
{
public static void main(String[] args)
{
int i, j, k;
//outer loop for the total rows which is 5 in this case for (i = 0 ; i<=5; i++)
{
int ch= 65;
//inner loop for the pattern of alphabets in till ‘i’ loop for (j = 0; j <=i ; j++)
{
System.out.print((char) (ch+j) + " ");
}
//inner loop for the pattern of alphabets in reverse order from ‘i’ loop for (k= i-1; k >=0; k--)
{
System.out.print((char) (ch+k) + " ");
}
System.out.println();
}
}
}

Output:

In the above example, if we observe each row of pattern, we need to print the alphabet first in the increasing order, i.e. A B and then in the reverse order, i.e. A B A. For this, we need 3 loops, 1st ?for? loop for the total number of rows. 2nd ?for? loop to print the alphabets in increasing order then the 3rd ?for? loop which remains inside the outer ‘i’ loop and prints the alphabets in the same line but in reverse order of ‘j’ loop.

Conclusion

The above example and their explanations clearly show how to make such patterns in Java. Though these patterns seem to be difficult in the starting, observing them deeply of how the repetition of pattern is happening in a single row and according to how many loops should be used, it becomes easy to do hands-on on this. Today also, in interviews of big companies, candidates are asked to write the logic of patterns of varying difficulty levels because this pattern making shows the basic logical and programming knowledge of an individual.

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