我的答案，请大家指教。

冯征sh

#1 Created at ... [Delete] [Delete and Lock User]

# 思路是循环从左边去除空格，然后循环从右边去除空格。

# 运算效率比递归高一些，从左边循环删除空格后，再从右边循环删除空格。

# 循环过程中不是空格立即退出循环。

def trim(s):    while length := len(s) >= 1:  # 海象运算符，测量字符串长度，>=1才循环        if s[:1] == ' ':  # 从左边判断是不是空格            s = s[1:]  # 去除左边的空格            length = len(s)  # 长度重新计算        else:  # 最左边判断不是空格的退出循环            break    while length := len(s) >= 1:  # 海象运算符，测量字符串长度，>=1才循环        if s[-1] == ' ':  # 从右边判断是不是空格的去除            s = s[:-1]  # 去除左边的空格            length = len(s)  # 长度重新计算        else:  # 最右边判断不是空格的退出循环            break    return s

冯征sh

#2 Created at ... [Delete] [Delete and Lock User]

def trim(s):    while length := len(s) >= 1:        if s[:1] == ' ':            s = s[1:]            length = len(s)        else:            break    while length := len(s) >= 1:        if s[-1] == ' ':            s = s[:-1]            length = len(s)        else:            break    return s

冯征sh

#3 Created at ... [Delete] [Delete and Lock User]

def trim(s):    while length := len(s) >= 1:        if s[:1] == ' ':            s = s[1:]            length = len(s)        else:            break    while length := len(s) >= 1:        if s[-1] == ' ':            s = s[:-1]            length = len(s)        else:            break    return s

冯征sh

#4 Created at ... [Delete] [Delete and Lock User]

def trim(s):

while length := len(s) >= 1:

if s[:1] == ' ':

s = s[1:]

length = len(s)

else:

break

while length := len(s) >= 1:

if s[-1] == ' ':

s = s[:-1]

length = len(s)

else:

break

return s

冯征sh

#5 Created at ... [Delete] [Delete and Lock User]

def trim(s):

while length := len(s) >= 1: # 海象运算符，测量字符串长度，>=1才循环

if s[:1] == ' ': # 从左边判断是不是空格

s = s[1:] # 去除左边的空格

length = len(s) # 长度重新计算

else: # 最左边判断不是空格的退出循环

break

while length := len(s) >= 1: # 海象运算符，测量字符串长度，>=1才循环

if s[-1] == ' ': # 从右边判断是不是空格的去除

s = s[:-1] # 去除左边的空格

length = len(s) # 长度重新计算

else: # 最右边判断不是空格的退出循环

break

return s

冯征sh

#6 Created at ... [Delete] [Delete and Lock User]

补充一下，python递归有996层级限制，最好别用递归解。

递归虽然简洁，但是运算效率低。

优先考虑循环解。