今天画一棵圣诞树比较应景。
当n=3时的圣诞树
如下图：
*
***
*****
###
当n=9时的圣诞树 如下图：
*
***
*****
***
*****
*******
*****
*******
*********
###
请编写一个函数，只接收一个数字n，代表树的层数，返回打印树形的字符串。
具体的绘画的规则是：
int(n/3)；题目难度：中等
题目来源：CodeWars：Merry Christmas
def solution(n: int) -> str:
# your code
assert solution(5) == " *\r\n ***\r\n*****\r\n ###"
assert solution(8) == "*\r\n ***\r\n *****\r\n ***\r\n *****\r\n*******\r\n ###"
def solution(n: int) -> str:
christmas = ''
pre_num = 1
space_num = n // 3 + 1
for _ in range(n // 3):
christmas += ' ' * space_num + pre_num * '*' + '\r\n' + ' ' * (space_num - 1) \
+ (pre_num + 2) * '*' + '\r\n' + ' ' * (space_num - 2) + (pre_num + 4) * '*' + '\r\n'
pre_num = pre_num + 2
space_num = space_num - 1
return christmas + ' ' * (n // 3) + "###"
assert solution(5) == " *\r\n ***\r\n*****\r\n ###"
assert solution(8) == " *\r\n ***\r\n *****\r\n ***\r\n *****\r\n*******\r\n ###"
def solution(n: int) -> str:
# 计算层数
floor = int(n / 3)
result = ''
# 遍历层数
for x in range(1, floor + 1):
# 定义每层的遍历开始值
start = 2 * x - 1
# 定义每层的遍历结束值
end = 2 * x + 4
# 开始生成每层的3行字符串，y是每行的*数量，每行的步长为2
for y in range(start, end, 2):
# 定义空格数量
num1 = int((floor * 2 + 3 - y) / 2)
# 空格数量 + *数量 + 回车换行
result += ' ' * num1 + '*' * y + '\r\n'
result += ' ' * floor + '###'
return result
def solution(n: int) -> str:
num = int(n / 3)
str = ''
for i in range(1, num + 1):
str += ' ' * (num-(i-1)+1) + '*' * (2 * i - 1) + "\r\n"
str += ' ' * (num-(i-1)) + '*' * (2 * i + 1) + "\r\n"
str += " " * (num-(i-1)-1) + "*" * (2 * i + 3) + "\r\n"
str += " " * num+"###"
return str
def solution(n: int) -> str:
# 层数
num = int(n / 3)
# 每层数据
data = []
# 遍历层数从第1层到num层
for j in range(1, num + 1):
# 第j层，每层3行
for i in range(j, 3 + j):
# 第i行 num+2-i个空格 2i-1个'*'号
line = ' ' * (num + 2 - i) + (2 * i - 1) * '*'
data.append(line)
# 最后加"###"
data.append(' ' * num + "###")
# 格式化处理
result = '\r\n'.join(data)
return result
assert solution(5) == " *\r\n ***\r\n*****\r\n ###"
assert solution(8) == " *\r\n ***\r\n *****\r\n ***\r\n *****\r\n*******\r\n ###"
def solution8(n):
# layer为圣诞树层数
layer=int(n/3)
y=1
while y<=layer:
for i in range((2*y-1),2*y+4,2):
ss='*'*i
print(ss.center(2*layer+3))
y+=1
print('###'.center(2*layer+3))
solution8(8)
def solution8(n):
# layer为圣诞树层数
layer=int(n/3)
y=1
christmas=''
while y<=layer:
for i in range((2*y-1),2*y+4,2):
ss='*'*i
christmas=(christmas+ss.center(2*layer+3)).rstrip()+'\r\n'
# print(ss.center(2*layer+3))
y+=1
return (christmas+'###'.center(2*layer+3)).rstrip()
assert solution8(5) == " *\r\n ***\r\n*****\r\n ###"
assert solution8(8) == " *\r\n ***\r\n *****\r\n ***\r\n *****\r\n*******\r\n ###"
def solution(n: int) -> str:
layer_num = n // 3 # 圣诞树的总层数
space_num_first = layer_num + 1 # 圣诞树第一层第一行 空格的数量 ，这个必须根据圣诞树总的行数来定
str = '' # 最后返回的 圣诞树字符串
for layer in range(1,layer_num+1): # 控制圣诞树的层数
star_num = 2 * layer - 1 # 圣诞树 每一层 第一行的星星数量
space_num = space_num_first - (layer - 1) # 圣诞树 每一层 第一行空格的数量
for n in range(1,4): # 控制圣诞树的每一层的行数
str = str+ ' ' * space_num + '*' * star_num + '\r\n'
star_num = star_num + 2 # 下一行比一行星星多2个
space_num = space_num -1 # 下一行的空格比上一行的少一个（左边才有空格）
str = str + ' ' * (space_num_first - 1) + '###' # 最后的 ### 前面的空格数量，比第一层第一行的空格数量少一个
return str
def solution(n: int) -> str:
level = int(n/3)
max_width = level * 2 + 3
_list = []
for cur_level in range(1,level+1):
for i in range(cur_level*2-1,cur_level * 2 + 4,2):
_list.append("{:>{width}}".format('*'*i,width=int((max_width-i)/2)+i))
_list.append("{:>{width}}".format("###", width=level + 3))
return "\r\n".join(_list)
def solution(n):
chrimas_tree = ""
# 获取圣诞树的层数
tree_floor = int(n/3)
if tree_floor<1:
return ""
for tree_floor_num in range(1, tree_floor+1):
# 根据圣诞树的层数计算出最顶上的空格数量,第一层space_num space_num-1 space_num-2 第二层为space_num-1 space_num-2 space_num-3
space_num = tree_floor + 1 - tree_floor_num + 1
chrimas_tree += " " * space_num + "*" * (2 * tree_floor_num - 1) + "\r\n" + " " * (space_num - 1) + "*" * (
2 * tree_floor_num + 1) + "\r\n" + " " * (space_num - 2) + "*" * (2 * tree_floor_num + 3)
if tree_floor_num == tree_floor:
chrimas_tree +="\r\n" + " "*(tree_floor)+"#"*3
else:
chrimas_tree += '\r\n'
# print(chrimas_tree)
return chrimas_tree
assert solution(2) == ""
assert solution(5) == " *\r\n ***\r\n*****\r\n ###"
assert solution(8) == " *\r\n ***\r\n *****\r\n ***\r\n *****\r\n*******\r\n ###"
def solution(n: int) -> str:
layer = 1
str_layer = ''
for i in range(0, int(n/3)):
tem_layer = layer
tem_str = (' '*(2+int(n/3)-1-i) +'*' * tem_layer +'\r\n'+' '*(1+int(n/3)-1-i) + '*' * (tem_layer+2)+'\r\n'+' '*(int(n/3)-1-i)+'*' * (tem_layer+4) + '\r\n')
str_layer += tem_str
layer += 2
return str_layer+' '*(n//3)+'###'
assert solution(5) == " *\r\n ***\r\n*****\r\n ###"
assert solution(8) == " *\r\n ***\r\n *****\r\n ***\r\n *****\r\n*******\r\n ###"