擅长:python、mysql、java
<p>根据原始字符串的模式和结果,我构造了这个算法。它的操作数最少。你知道吗</p>
<pre><code>str = 'Strings'
lens = len(str)
lensh = int(lens/2)
nstr = ''
for i in range(lensh):
nstr = nstr + str[lens - i - 1] + str[i]
if ((lens % 2) == 1):
nstr = nstr + str[lensh]
print(nstr)
</code></pre>