練習(xí)
楊輝三角定義如下:
1
/ \
1 1
/ \ / \
1 2 1
/ \ / \ / \
1 3 3 1
/ \ / \ / \ / \
1 4 6 4 1
/ \ / \ / \ / \ / \
1 5 10 10 5 1
把每一行看做一個(gè)list,試寫(xiě)一個(gè)generator,不斷輸出下一行的list:
期待輸出: # [1] # [1, 1] # [1, 2, 1] # [1, 3, 3, 1] # [1, 4, 6, 4, 1] # [1, 5, 10, 10, 5, 1] # [1, 6, 15, 20, 15, 6, 1] # [1, 7, 21, 35, 35, 21, 7, 1] # [1, 8, 28, 56, 70, 56, 28, 8, 1] # [1, 9, 36, 84, 126, 126, 84, 36, 9, 1]
思路:找規(guī)律
第一行:[1]
第二行:[0, 1] + [1, 0] = [1, 1]
第三行:[0, 1, 1] +?[1, 1, 0] = [1, 2, 1]
第四行:[0, 1, 2, 1] +?[1, 2, 1, 0] = [1, 3, 3, 1]
第五行:[0, 1, 3, 3, 1] +?[1, 3, 3, 1, 0] = [1, 4, 6, 4, 1]
…… ……
可以看出,每一行的List都是由上一行的List分別在頭、尾加上一個(gè)0之后新生成的兩個(gè)List對(duì)應(yīng)位置元素之和,求兩個(gè)List對(duì)應(yīng)位置元素之和可以利用 zip()函數(shù)
即:L=[ sum[ i ] for i in zip( [0]+L, L+[0] ) ]
?
|
? 代碼:
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import math
def triangles():
L = [1]
while True:
yield L
zipped=list(zip([0]+L, L+[0]))
L = [sum(i) for i in zip([0]+L, L+[0])]
# 期待輸出:
# [1]
# [1, 1]
# [1, 2, 1]
# [1, 3, 3, 1]
# [1, 4, 6, 4, 1]
# [1, 5, 10, 10, 5, 1]
# [1, 6, 15, 20, 15, 6, 1]
# [1, 7, 21, 35, 35, 21, 7, 1]
# [1, 8, 28, 56, 70, 56, 28, 8, 1]
# [1, 9, 36, 84, 126, 126, 84, 36, 9, 1]
n = 0
results = []
for t in triangles():
print(t)
results.append(t)
n = n + 1
if n == 10:
break
if results == [
[1],
[1, 1],
[1, 2, 1],
[1, 3, 3, 1],
[1, 4, 6, 4, 1],
[1, 5, 10, 10, 5, 1],
[1, 6, 15, 20, 15, 6, 1],
[1, 7, 21, 35, 35, 21, 7, 1],
[1, 8, 28, 56, 70, 56, 28, 8, 1],
[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]
]:
print('測(cè)試通過(guò)!')
else:
print('測(cè)試失敗!')
?
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