标签：bre [1] pos 逻辑 numpy 切片 浅拷贝 copy 可视化
The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
P A H N A P L S I I G Y I R
And then read line by line: "PAHNAPLSIIGYIR"
Write the code that will take a string and make this conversion given a number of rows:
string convert(string text, int nRows);
convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".
代码如下：
1 import copy 2 class Solution(object): 3 def convert(self, s, numRows): 4 """ 5 :type s: str 6 :type numRows: int 7 :rtype: str 8 """ 9 def postion_next(x, y): 10 11 if x == 0: 12 return x + 1, y 13 elif x == r - 1: 14 return x - 1, y + 1 15 elif mat[x - 1][y] == ‘ ‘: 16 return x - 1, y + 1 17 else: 18 return x + 1, y 19 20 r = numRows 21 if s == "": 22 return "" 23 if r == 1: 24 return s 25 c = (int(len(s) / (2 * r - 2)) + 1) * (r - 1) 26 line = [] 27 mat = [] 28 for i in range(c): 29 line.append(‘ ‘) 30 for i in range(r): 31 mat.append(copy.deepcopy(line)) 32 l = len(s) 33 postion_s = 0 34 postion_m = [0, 0] 35 while True: 36 # print(postion_m) 37 if (postion_s) == l: 38 break 39 mat[postion_m[0]][postion_m[1]] = s[postion_s] 40 postion_s += 1 41 postion_m = postion_next(postion_m[0], postion_m[1]) 42 43 # for i in range(r): 44 # for j in range(c): 45 # print(mat[i][j], end=‘‘) 46 # if j == c - 1: 47 # print(‘\n‘, end=‘‘) 48 res = [] 49 for i in range(r): 50 res += mat[i] 51 res = ‘‘.join(res) 52 # print(‘‘.join(res.split(‘ ‘))) 53 return ‘‘.join(res.split(‘ ‘))
结果：超时，不过逻辑没问题，注释部分是可视化输出z字形打印。
总结：
1.原生python矩阵生成操作好麻烦，用numpy的话np.array(list(‘ ‘*9)).reshape(3,3)就可以生成3*3的字符型矩阵
2.复习到了copy库的知识，如果不进行deepcopy的话下面的代码中mat的全部r层会浅拷贝line。
1 line = [] 2 mat = [] 3 for i in range(c): 4 line.append(‘ ‘) 5 for i in range(r): 6 mat.append(copy.deepcopy(line))
3.复习了str.split()的str切片剔除分隔符为list操作，顺便str.strip()开头结尾剥离操作，‘‘.join(list)列表合并为str操作。
标签：bre [1] pos 逻辑 numpy 切片 浅拷贝 copy 可视化
原文地址：http://www.cnblogs.com/hellcat/p/6837419.html
