标签:count down Conversion 字形 res up ZigZag numRows str
ZigZag Conversion
将一个给定字符串根据给定的行数,以从上往下、从左到右进行 Z 字形排列。
题目来源:力扣(LeetCode)链接:https://leetcode-cn.com/problems/zigzag-conversion
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 s, int numRows);
Example 1:
Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"
Example 2:
Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:
P I N
A L S I G
Y A H R
P I
python3:
class Solution:
def convert(self, s: str, numRows: int) -> str:
str_len = len(s)
convert_list = [ [''] * str_len for i in range(numRows)]
i = 0
j = 0
count = 0
up_to_down = True
while count < str_len:
convert_list[j][i] = s[count]
count = count + 1
if up_to_down:
j = j + 1
else:
j = j - 1
i = i + 1
if j >= numRows:
i = i + 1
j = j - 2
up_to_down = False
elif j < 0:
j = j + 2
up_to_down = True
res_str = ''
for lists in convert_list:
for l in lists:
res_str = res_str + l
return res_str
改进:
class Solution:
def convert(self, s: str, numRows: int) -> str:
result_list = [''] * numRows
i = 0
j = 0
count = 0
up_to_down = True
str_len = len(s)
while count < str_len:
result_list[j] = result_list[j] + s[count]
count = count + 1
if up_to_down:
j = j + 1
else:
j = j - 1
i = i + 1
if j >= numRows:
i = i + 1
j = j - 2
up_to_down = False
elif j < 0:
j = j + 2
up_to_down = True
res_str = ''
for res in result_list:
res_str = res_str + res
return res_str
标签:count,down,Conversion,字形,res,up,ZigZag,numRows,str 来源: https://blog.csdn.net/qq_15706279/article/details/100585359
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