DAY 011--杨辉三角
011题目
杨辉三角定义如下:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
问题:给出制定的值,输出直到该行的杨辉三角
流程分析:
1、找规律:
- 每一行的开头和结尾都是一个1
- 每行的中间的值都是前一行每个list[i]+list[i+1]的值
2、程序流程
- 写一个杨辉三角的函数,起始值b=[1]
- 一直循环,yield b(yield目前还不是很明白,以后复习到那里再回来写心得)
- b=[1]+[b[i]+b[i+1] for i in range(len(i)-1)]+[1]
代码分析:
def triangles(): b = [1] while True: yield b b = [1] + [b[i] + b[i+1] for i in range(len(b)-1)] + [1] n = 0 for t in triangles(): print(t) n = n + 1 if n == 20:#前20行的杨辉三角 break #运行结果 [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] [1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1] [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1] [1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1] [1, 13, 78, 286, 715, 1287, 1716, 1716, 1287, 715, 286, 78, 13, 1] [1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1] [1, 15, 105, 455, 1365, 3003, 5005, 6435, 6435, 5005, 3003, 1365, 455, 105, 15, 1] [1, 16, 120, 560, 1820, 4368, 8008, 11440, 12870, 11440, 8008, 4368, 1820, 560, 120, 16, 1] [1, 17, 136, 680, 2380, 6188, 12376, 19448, 24310, 24310, 19448, 12376, 6188, 2380, 680, 136, 17, 1] [1, 18, 153, 816, 3060, 8568, 18564, 31824, 43758, 48620, 43758, 31824, 18564, 8568, 3060, 816, 153, 18, 1] [1, 19, 171, 969, 3876, 11628, 27132, 50388, 75582, 92378, 92378, 75582, 50388, 27132, 11628, 3876, 969, 171, 19, 1]
题目反思:
1、找规律的时候主要难点在于能够找出,下一列中间的值为上一列的:list[i]+list[i+1] for i in range(len(i)-1)
2、yield方法目前还不是很熟悉,目前只知道这是生成器,等过几天搞明白后再来回味这道题
新学知识点:
1、杨辉三角的定义,以及杨辉三角如何用代码实现
2、yield知识点空缺
Mark on 2018.04.13