替换空格

╰半夏微凉° 2022-06-08 02:18 274阅读 0赞

> 牛客网 《剑指offer》
> 
> 时间限制：`1秒` 空间限制：`32768K` 热度指数：`378402`  
> 本题知识点： 字符串

**题目描述**

请实现一个函数，将一个字符串中的空格替换成“%20”。例如，当字符串为We Are Happy.则经过替换之后的字符串为We%20Are%20Happy。

**解题思路**

![这里写图片描述][SouthEast]

> 参考别人的Python实现  
> [https://github.com/Jack-Lee-Hiter/AlgorithmsByPython/blob/master/Target%20Offer/%E6%9B%BF%E6%8D%A2%E7%A9%BA%E6%A0%BC.py][https_github.com_Jack-Lee-Hiter_AlgorithmsByPython_blob_master_Target_20Offer_E6_9B_BF_E6_8D_A2_E7_A9_BA_E6_A0_BC.py]

''' 请实现一个函数，将一个字符串中的空格替换成“%20”。 例如，当字符串为We Are Happy.则经过替换之后的字符串为We%20Are%20Happy。 '''
    
    # -*- coding:utf-8 -*-
    class Solution:
        # s 源字符串
    
        # 创建新的字符串进行替换
        def replaceSpace1(self, s):
            tempstr = ''
            if type(s) != str:
                return
            for c in s:
                if c == ' ':
                    tempstr += '%20'
                else:
                    tempstr += c
            return tempstr
    
        # 简单代码替换
        # 在Python中str类型是不可变的类型, 使用replace语句会生成一个新的str, 原始的s还是带空格的str变量
        def replaceSpace2(self, s):
            if type(s) != str:
                return
            return s.replace(' ', '%20')
    
        # 书中给的思路
        # 判断输入类型的时候，isinstance必须首先判断，因为如果输入为integer的话，没有len，就会直接报错
        def replaceSpace3(self, s):
            if not isinstance(s,str) or len(s) <= 0 or s == None:
                return ""
            spaceNum = 0
            for i in s:
                if i == " ":
                    spaceNum += 1
    
            newStrLen = len(s) + spaceNum * 2
            newStr = newStrLen * [None]
            indexOfOriginal, indexOfNew = len(s) - 1, newStrLen - 1
            while indexOfNew >= 0 and indexOfNew >= indexOfOriginal:
                if s[indexOfOriginal] == ' ':
                    newStr[indexOfNew-2:indexOfNew+1] = ['%', '2', '0']
                    indexOfNew -= 3
                    indexOfOriginal -= 1
                else:
                    newStr[indexOfNew] = s[indexOfOriginal]
                    indexOfNew -= 1
                    indexOfOriginal -= 1
            return "".join(newStr)
    
    
    s = 'we are happy'
    test = Solution()
    print(test.replaceSpace1(s))
    print(test.replaceSpace2(s))
    print(test.replaceSpace3(s))

[SouthEast]: /images/20220608/54b2d7489f844dd4b11b571bce56cfb5.png
[https_github.com_Jack-Lee-Hiter_AlgorithmsByPython_blob_master_Target_20Offer_E6_9B_BF_E6_8D_A2_E7_A9_BA_E6_A0_BC.py]: https://github.com/Jack-Lee-Hiter/AlgorithmsByPython/blob/master/Target%20Offer/%E6%9B%BF%E6%8D%A2%E7%A9%BA%E6%A0%BC.py