05. 替换空格

题目：请实现一个函数，将一个字符串中的空格替换成“%20”

=======================================1=========================================

def replaceSpace(str1):
    """
    有额外的空间;从后面逐个往前替换，不可从前往后，因为字符串的长度越来越长
    """
    length = len(str1)
    li = range(length)[::-1]
    for i in li:
        if str1[i]== " ":
            str1left = str1[:i]
            str1right = str1[i+1:]
            str1 = str1left + "%20" + str1right
    return str1

# 测试
for stri in [" sfdvd fvd  dgfd  sfr ", "sdfsgef", "", " ", "   "]:
    print(stri, "=>", replaceSpace(stri))

 

sfdvd fvd  dgfd  sfr  => %20sfdvd%20fvd%20%20dgfd%20%20sfr%20
sdfsgef => sdfsgef
 => 
  => %20
    => %20%20%20

=======================================2=========================================

def replaceSpace2(str1):
    """
    有额外的空间
    """
    res = ""
    for i in range(len(str1)):
        if str1[i]== " ":
            res = res + "%20"
        else:
            res = res + str1[i]
    return res

# 测试
for stri in [" sfdvd fvd  dgfd  sfr ", "sdfsgef", "", " ", "   "]:
    print(stri, "=>", replaceSpace2(stri))

sfdvd fvd  dgfd  sfr  => %20sfdvd%20fvd%20%20dgfd%20%20sfr%20
sdfsgef => sdfsgef
 => 
  => %20
    => %20%20%20

==========================================3======================================

def replaceSpace3(str1):
    """
    提供的空间有限; 由于python中字符串属于不可修改的类型，
    所以不可通过改变字符串的方法,但是可以将字符串变成字符列表,再进行修改
    """
    str1 = list(str1)
    length = len(str1)
    if length == 0:
        return "".join(str1)
    # 统计有多少个空格
    num = 0
    for i in range(length):
        if str1[i]== " ":
            num += 1
    
    if num == 0:
        return "".join(str1)
    
    str1 = str1 + list(" " * (2 * num))
    new_length = len(str1)
    new_index = new_length - 1
    index = length - 1
    
    while(index >= 0):
        if(str1[index] == " "):
            str1[new_index] = "0"
            str1[new_index - 1] = "2"
            str1[new_index - 2] = "%"
            new_index = new_index - 3
        else:
            str1[new_index] = str1[index]
            new_index = new_index - 1
        index = index - 1
    return "".join(str1)

# 测试
for stri in [" sfdvd fvd  dgfd  sfr ", "sdfsgef", "", " ", "   "]:
    print(stri, "=>", replaceSpace3(stri))

sfdvd fvd  dgfd  sfr  => %20sfdvd%20fvd%20%20dgfd%20%20sfr%20
sdfsgef => sdfsgef
 => 
  => %20
    => %20%20%20

========================================4========================================

def replaceSpace4(str1): # 这相当于作弊行为
    return str1.replace(' ','%20')

# 测试
for stri in [" sfdvd fvd  dgfd  sfr ", "sdfsgef", "", " ", "   "]:
    print(stri, "=>", replaceSpace4(stri))

 

sfdvd fvd  dgfd  sfr  => %20sfdvd%20fvd%20%20dgfd%20%20sfr%20
sdfsgef => sdfsgef
 => 
  => %20
    => %20%20%20

=========================================5=======================================

def replaceSpace5(str1): # 另一种作弊行为,但是让我重新认识了split()函数
    return '%20'.join(str1.split(' '))

# 测试
for stri in [" sfdvd fvd  dgfd  sfr ", "sdfsgef", "", " ", "   "]:
    print(stri, "=>", replaceSpace5(stri))

 

sfdvd fvd  dgfd  sfr  => %20sfdvd%20fvd%20%20dgfd%20%20sfr%20
sdfsgef => sdfsgef
 => 
  => %20
    => %20%20%20