空格的替换
2019-12-19 本文已影响0人
而立之年的技术控
微信图片_20191219180714.jpg
class Solution:
# s 源字符串
def replaceSpace(self, s):
# write code here
if not s:
return ''
str1 = list(s)
count = str1.count(' ')
str2 = len(str1)*[' '] + [' ']*count*2
p1 = len(str1) - 1
p2 = len(str2) - 1
while p1 >= 0:
if str1[p1] == ' ':
str2[p2] = '0'
str2[p2-1] = '2'
str2[p2-2] = '%'
p1 -= 1
p2 -= 3
else:
str2[p2] = str1[p1]
p1 -= 1
p2 -= 1
return ''.join(str2)
