【解決方法】このパターンを Python で記述するにはどうすればよいですか?

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指定された行数について、Python で次のパターンを出力します。
N = 5 のパターン
1 2 3 4 5
11 12 13 14 15
21 22 23 24 25
16 17 18 19 20
6 7 8 9 10

私が試したこと：

Python

n = int(input())
upper = int(n/2)+1
lower = n - upper
start_i_lower = 0

for i in range(0, upper):
  if i != 0:
    i = i + i
  start = n*i + 1
  end = start + n

  for j in range(start, end):
    print(j, end =" ")
  start_i_lower = i
  print()

start_i_lower -= 1

for i in range(0, lower):
  if(n%2 == 0):
    start_i_lower -= 1

  start = n*(start_i_lower-1) + 1
  end = start + n

  for j in range(1, n+1):
        print(j+n, end =" ")

  print()

解決策 1
解決策 7
解決策 5
解決策 6

解決策 1

これを試して：

Python

n = 5
h = 0
d = int(n/2)+1
for r in range(0, n):
	s=''
	for c in range(1, n+1):
		if r<d:
			h = r*2
		elif r==d:
			h = d
		else:
			h = int(d/3)
		s+=str(h*n+c) + ' '
	print(s)

結果：

1 2 3 4 5 
11 12 13 14 15 
21 22 23 24 25 
16 17 18 19 20 
6 7 8 9 10

解決策 7

n = int(input())
if n % 2 == 0:
    n1 = int(n/2)
    n2 = n-n1
else:
    n1 = int(n/2)+1
    n2 = n-n1

x = 1
i = 1
i1 = 1

#Upper pattern
while i <= n1:
    for j in range(1, n+1):
        print(x, end=" ")
        x = x+1
    print()
    if i == n1:
        break
    x = x+n
    i = i+1


#Lowe Pattern
if n % 2 == 0:
    while i1 <= n2:
        for k in range(1, n+1):
            print(x, end=" ")
            x = x+1
        x=x-(2*n)
        print()
        i1 = i1+1
        x = x-n
        y=y+1

else:
    while i1 <= n2:
        x = x-(2*n)
        for k in range(1, n+1):
            print(x, end=" ")
            x = x+1
        x=x-n
        print()
        i1 = i1+1

解決策 5

より最適化されたバージョンは次のようになります。

Python

n = 5
h = 0
d = int(n/2)+1
for r in range(0, n):
	s=''
	if r<d:
	    h = r*2
	elif r==d:
		h = d
	else:
	    h = int(d/3)
	for c in range(1, n+1):
		s+=str(h*n+c) + ' '
	print(s)

解決策 6

n = int(input( ))
P = 1
for i in range(1, n + 1):
    for j in range(P, P + n):
        print(j, end=" ")
    print()
    if i == ((n + 1)//2):
        if (n % 2) != 0:
            P = n*(n - 2) + 1
        else:
            P = n * (n - 1) + 1
    elif i > ((n + 1) // 2):
        P = P - 2 * n
    else:
        P = P + 2 * n