要求
输入: s = "we are tester"
输出:we%20are%20tester
index() 函数语法
# index() 方法检测字符串中是否包含子字符串 str
fruits = ['apple', 'banana', 'cherry']
r = fruits.index("cherry")
print(r) # 2
替换函数
# -*- coding: utf-8 -*-
def replaceSpace(s):
"""
把字符串 s 中的每个空格替换成"%20
:param s: 字符串
:return:
"""
li = [] # 定义空列表
# 遍历
for i in s:
li.append(i)
# print(li) ['w', 'e', ' ', 'a', 'r', 'e', ' ', 't', 'e', 's', 't', 'e', 'r']
for i in li:
if i == ' ':
li[li.index(i)] = '%20'
return ''.join(li)
if __name__ == '__main__':
s = "we are tester"
r = replaceSpace(s)
print(r) # we%20are%20tester
方法二
python里面有个replace方法可以直接替换字符串,
但是一般面试官对这个回答不太满意,因为.......
s = "We are tester"
print(s.replace(" ", "%20")) # We%20are%20tester
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