任务是为给定的整数值打印示例中所示的模式。
目的不是仅仅打印这种模式,而是要学习解决此类问题的最佳方法,因为在编码考试和工作面试中经常会问这些问题。
例子:
Input: N = 4
Output:
4 4 4 4 4 4 4
4 3 3 3 3 3 4
4 3 2 2 2 3 4
4 3 2 1 2 3 4
4 3 2 2 2 3 4
4 3 3 3 3 3 4
4 4 4 4 4 4 4
Input: N = 7
Output:
7 7 7 7 7 7 7 7 7 7 7 7 7
7 6 6 6 6 6 6 6 6 6 6 6 7
7 6 5 5 5 5 5 5 5 5 5 6 7
7 6 5 4 4 4 4 4 4 4 5 6 7
7 6 5 4 3 3 3 3 3 4 5 6 7
7 6 5 4 3 2 2 2 3 4 5 6 7
7 6 5 4 3 2 1 2 3 4 5 6 7
7 6 5 4 3 2 2 2 3 4 5 6 7
7 6 5 4 3 3 3 3 3 4 5 6 7
7 6 5 4 4 4 4 4 4 4 5 6 7
7 6 5 5 5 5 5 5 5 5 5 6 7
7 6 6 6 6 6 6 6 6 6 6 6 7
7 7 7 7 7 7 7 7 7 7 7 7 7
对于本教程,使用了N = 4的示例。
- 步骤1:首先,分析图案的任何对称线。在这里,我们的图案在垂直和水平方向上都是对称的,因此绘制这样的对称线,
在将图案分成几部分后,首先尝试仅绘制左上部分,即A部分。如果没有任何对称线,请跳至步骤2。
- 第2步:现在将每个单元格(即元素)与行号和列号相关联(通常分别由i和j表示),就像
从现在开始,一个单元格用C(i,j)表示,其行号和列号为。
- 步骤3:在此步骤中,尝试查找C(i,j)的值与i和/或j之间的关系。通常,C的值取决于N的值以及i和j的相对值。详细说明,
–在第一行中,每个元素都相同,即4(= N)。因此,它没有太大帮助。
–在第二行中,对于i> = j,可以看到C从4减少到3,然后对于下一个i = j,它保持3,然后对于所有j的值,对于下一个i = j,它保持2,可以看到C从4减少到1。因此,在每一行中,C从N开始,递减1直到i> = j,然后变为常数。这里的C值取决于i和j之间的较小者,公式可以是:
C(i, j) = N - min(i, j) + 1
因此,我们的方法应该是:
C++
#include
using namespace std; int main() { int N = 4, i, j, min; cout << "Value of N: " << N << endl; for (i = 1; i <= N; i++) { for (j = 1; j <= N; j++) { min = i < j ? i : j; cout << N - min + 1; } cout << endl; } return 0; }
Java
// Java program to find triplets // a[i]>a[j]>a[k] and i
Python3
# Python3 program to find triplets # a[i]>a[j]>a[k] and i
C#
// C# program to find triplets // a[i]>a[j]>a[k] and i
C++
#include
using namespace std; int main() { int N = 4, i, j, min; cout << "Value of N: " << N << endl; for (i = 1; i <= N; i++) { for (j = 1; j <= N; j++) { min = i < j ? i : j; cout << N - min + 1; } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; cout << N - min + 1; } cout << endl; } return 0; }
Java
class GFG { public static void main(String[] args) { int N = 4, i, j, min; System.out.println("Value of N: " + N); for (i = 1; i <= N; i++) { for (j = 1; j <= N; j++) { min = i < j ? i : j; System.out.print(N - min + 1); } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; System.out.print(N - min + 1); } System.out.println(); } } } // This code is contributed by PrinciRaj1992
Python3
N = 4; print("Value of N: ", N); for i in range(1, N + 1): for j in range(1, N + 1): min = i if i < j else j; print(N - min + 1, end = ""); for j in range(N - 1, 0, -1): min = i if i < j else j; print(N - min + 1, end = ""); print(); # This code is contributed by Rajput-Ji
C#
using System; class GFG { public static void Main(String[] args) { int N = 4, i, j, min; Console.WriteLine("Value of N: " + N); for (i = 1; i <= N; i++) { for (j = 1; j <= N; j++) { min = i < j ? i : j; Console.Write(N - min + 1); } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; Console.Write(N - min + 1); } Console.WriteLine(); } } } // This code is contributed by Rajput-Ji
C++
#include
using namespace std; int main() { int N = 4, i, j, min; cout << "Value of N: " << N << endl; for (i = 1; i <= N; i++) { for (j = 1; j <= N; j++) { min = i < j ? i : j; cout << N - min + 1; } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; cout << N - min + 1; } cout << endl; } for (i = N - 1; i >= 1; i--) { for (j = 1; j <= N; j++) { min = i < j ? i : j; cout << N - min + 1; } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; cout << N - min + 1; } cout << endl; } return 0; }
Java
// Java implementation of the approach class GFG { public static void main(String[] args) { int N = 4, i, j, min; System.out.println("Value of N: " + N); for (i = 1; i <= N; i++) { for (j = 1; j <= N; j++) { min = i < j ? i : j; System.out.print(N - min + 1); } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; System.out.print(N - min + 1); } System.out.println(); } for (i = N - 1; i >= 1; i--) { for (j = 1; j <= N; j++) { min = i < j ? i : j; System.out.print(N - min + 1); } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; System.out.print(N - min + 1); } System.out.println(); } } } // This code is contributed by 29AjayKumar
Python3
# Python3 implementation of the approach N = 4 print("Value of N: ", N) for i in range(1, N + 1): for j in range(1, N + 1): min = i if i < j else j print(N - min + 1, end = " ") for j in range(N - 1, 0, -1): min = i if i < j else j print(N - min + 1, end = " ") print() for i in range(N - 1, 0, -1): for j in range(1, N + 1): min = i if i < j else j print(N - min + 1, end = " ") for j in range(N - 1, 0, -1): min = i if i < j else j print(N - min + 1, end = " ") print() # This code is contributed by sai prakash
C#
// C# implementation of the approach using System; class GFG { public static void Main(String[] args) { int N = 4, i, j, min; Console.WriteLine("Value of N: " + N); for (i = 1; i <= N; i++) { for (j = 1; j <= N; j++) { min = i < j ? i : j; Console.Write(N - min + 1); } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; Console.Write(N - min + 1); } Console.WriteLine(); } for (i = N - 1; i >= 1; i--) { for (j = 1; j <= N; j++) { min = i < j ? i : j; Console.Write(N - min + 1); } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; Console.Write(N - min + 1); } Console.WriteLine(); } } } // This code is contributed by Rajput-Ji
输出:Value of N: 4 4444 4333 4322 4321
如果图案没有任何对称性,并且现在完成了,那么工作就完成了。但是对于对称的图案,它们仍然不完整,需要执行步骤4。
- 步骤4:现在,在图片中包括图案的B部分,并将这些元素与列号相关联。但是不要按顺序分配列号,而是给它们分配与A部分中为其镜像列分配的列号相同的列号。
- 步骤5:现在更改代码以添加B部分。为此,只需重新运行j = n-1到j = 1的内部循环。
C++
#include
using namespace std; int main() { int N = 4, i, j, min; cout << "Value of N: " << N << endl; for (i = 1; i <= N; i++) { for (j = 1; j <= N; j++) { min = i < j ? i : j; cout << N - min + 1; } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; cout << N - min + 1; } cout << endl; } return 0; } Java
class GFG { public static void main(String[] args) { int N = 4, i, j, min; System.out.println("Value of N: " + N); for (i = 1; i <= N; i++) { for (j = 1; j <= N; j++) { min = i < j ? i : j; System.out.print(N - min + 1); } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; System.out.print(N - min + 1); } System.out.println(); } } } // This code is contributed by PrinciRaj1992
Python3
N = 4; print("Value of N: ", N); for i in range(1, N + 1): for j in range(1, N + 1): min = i if i < j else j; print(N - min + 1, end = ""); for j in range(N - 1, 0, -1): min = i if i < j else j; print(N - min + 1, end = ""); print(); # This code is contributed by Rajput-Ji
C#
using System; class GFG { public static void Main(String[] args) { int N = 4, i, j, min; Console.WriteLine("Value of N: " + N); for (i = 1; i <= N; i++) { for (j = 1; j <= N; j++) { min = i < j ? i : j; Console.Write(N - min + 1); } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; Console.Write(N - min + 1); } Console.WriteLine(); } } } // This code is contributed by Rajput-Ji
输出:Value of N: 4 4444444 4333334 4322234 4321234
- 步骤6:现在,在图片中包括图案的C和D部分,并将这些元素与行号相关联,方法与在步骤4中分配列号的方式相同。
- 步骤7:现在,以与步骤5中相同的方式,更改代码以追加C和D部分。为此,只需重新运行外循环,即可将i = n-1转换为i = 1。
C++
#include
using namespace std; int main() { int N = 4, i, j, min; cout << "Value of N: " << N << endl; for (i = 1; i <= N; i++) { for (j = 1; j <= N; j++) { min = i < j ? i : j; cout << N - min + 1; } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; cout << N - min + 1; } cout << endl; } for (i = N - 1; i >= 1; i--) { for (j = 1; j <= N; j++) { min = i < j ? i : j; cout << N - min + 1; } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; cout << N - min + 1; } cout << endl; } return 0; } Java
// Java implementation of the approach class GFG { public static void main(String[] args) { int N = 4, i, j, min; System.out.println("Value of N: " + N); for (i = 1; i <= N; i++) { for (j = 1; j <= N; j++) { min = i < j ? i : j; System.out.print(N - min + 1); } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; System.out.print(N - min + 1); } System.out.println(); } for (i = N - 1; i >= 1; i--) { for (j = 1; j <= N; j++) { min = i < j ? i : j; System.out.print(N - min + 1); } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; System.out.print(N - min + 1); } System.out.println(); } } } // This code is contributed by 29AjayKumar
Python3
# Python3 implementation of the approach N = 4 print("Value of N: ", N) for i in range(1, N + 1): for j in range(1, N + 1): min = i if i < j else j print(N - min + 1, end = " ") for j in range(N - 1, 0, -1): min = i if i < j else j print(N - min + 1, end = " ") print() for i in range(N - 1, 0, -1): for j in range(1, N + 1): min = i if i < j else j print(N - min + 1, end = " ") for j in range(N - 1, 0, -1): min = i if i < j else j print(N - min + 1, end = " ") print() # This code is contributed by sai prakash
C#
// C# implementation of the approach using System; class GFG { public static void Main(String[] args) { int N = 4, i, j, min; Console.WriteLine("Value of N: " + N); for (i = 1; i <= N; i++) { for (j = 1; j <= N; j++) { min = i < j ? i : j; Console.Write(N - min + 1); } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; Console.Write(N - min + 1); } Console.WriteLine(); } for (i = N - 1; i >= 1; i--) { for (j = 1; j <= N; j++) { min = i < j ? i : j; Console.Write(N - min + 1); } for (j = N - 1; j >= 1; j--) { min = i < j ? i : j; Console.Write(N - min + 1); } Console.WriteLine(); } } } // This code is contributed by Rajput-Ji
输出:Value of N: 4 4444444 4333334 4322234 4321234 4322234 4333334 4444444