leetcode_剑指 Offer 05. 替换空格

请实现一个函数,把字符串 s 中的每个空格替换成"%20"。

 

示例 1:

输入:s = "We are happy."
输出:"We%20are%20happy."
 

限制:

0 <= s 的长度 <= 10000

作者:Krahets
链接:https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/50ywkd/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。
class Solution:
    def replaceSpace(self, s: str) -> str:
        t=''
        for x in s:
            if x==' ':
                t+='%20'
            else:
                t+=x
        return t
原文地址:https://www.cnblogs.com/hqzxwm/p/14377886.html