Leetcode ZigZag Conversion
Leetcode ZigZag Conversion
丹
发布于 2018-09-03 18:16:04
发布于 2018-09-03 18:16:04
代码可运行
运行总次数：0
代码可运行
题目：
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 RAnd 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解法：
自己的解法如下：
class Solution:
def convert(self, s, numRows):
"""
:type s: str
:type numRows: int
:rtype: str
"""
res=[]
n=len(s)
if n<=0:
return ""
if numRows==1:
return s
tmp=[]
for i in s:
tmp.append(i)
index = 0
while len(tmp)>0:
if index==0:
cur=tmp[:numRows]
while len(cur)<numRows:
cur.append('')
res.append(cur)
index=numRows-2
tmp=tmp[numRows:]
else:
cur=['']*numRows
cur[index]=tmp[0]
res.append(cur)
tmp = tmp[1:]
index-=1
res_str=''
for i in range(numRows):
for j in range(len(res)):
if res[j][i]:
res_str+=res[j][i]
return res_str参考解法：ZigZag Conversion
class Solution:
def convert(self, s, numRows):
"""
:type s: str
:type numRows: int
:rtype: str
"""
if numRows == 1: return s
temp = ['' for i in range(numRows)]
index = -1
step = 1
for i in range(len(s)):
index += step
if index == numRows:
index -= 2
step = -1
if index == -1:
index = 1
step = 1
temp[index] += s[i]
return "".join(temp)只用了一个数组，并且将按照最终读取的顺序将字符存到对应位置。
本文参与 腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。
原始发表：2018年08月12日，如有侵权请联系 cloudcommunity@tencent.com 删除
评论
登录后参与评论
推荐阅读
目录
腾讯云开发者
