我玩耍了,這裏是結果
nonechar = 'N'
spacechar = '_'
solution = [[6], [0, 7], [None, 3, None, 8], [2, 5, None, 9], [1, None, 4, None, None, None], [None, None, None, 4],[None, 3]]
for i in range(1, len(solution)):
for j in range(len(solution[i-1])):
if (solution[i-1][j] == None):
solution[i].insert(2*j, None)
solution[i].insert(2*j+1, None)
N = len(solution[-1]) * 2 - 1
offset = (N - 1)/2
spacing = 0
for i in range(len(solution)):
line = spacechar * int(offset)
for j in range(len(solution[i])):
if (solution[i][j] == None):
line += nonechar
else:
line += str(solution[i][j])
if (j != len(solution[i]) - 1):
line += spacechar * int(spacing)
line += spacechar * int(offset)
print(line)
spacing = offset
offset = (offset - 1)/2
我基本上做的是用缺少的數據填充解決方案列表,以便每個下一個子列表的值比前一個值多兩倍。對於i
第一個子列表中的每個j
第012個元素,都有[i+1][2*j]
和[i+1][2*j+1]
下的值。然後我使用ASCII藝術打印出結果,計算出所需的偏移量和間距。這裏的限制是,你只能使用數字0-9,以免搞亂我的樹。你必須找出解決方法,你自己:)
哦,是的。輸出看起來像這樣(隨意更改缺少值和空格的字符):
_______________________________________________________________6_______________________________________________________________
_______________________________0_______________________________________________________________7_______________________________
_______________N_______________________________3_______________________________N_______________________________8_______________
_______N_______________N_______________2_______________5_______________N_______________N_______________N_______________9_______
___N_______N_______N_______N_______1_______N_______4_______N_______N_______N_______N_______N_______N_______N_______N_______N___
_N___N___N___N___N___N___N___N___N___N___N___N___N___4___N___N___N___N___N___N___N___N___N___N___N___N___N___N___N___N___N___N_
N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_3_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N_N