擅长:python、mysql、java
<p>严格的解决方案:</p>
<pre><code>s = str(1234567890)
def processing(s):
i=0
while i < len(s):
yield s[i:i+2]
i = i+2
[x for x in processing(s)]
</code></pre>
<p>输出:</p>
<p>['12'、'34'、'56'、'78'、'90']</p>
<p>如果希望它是一个字符串,用空格分隔整数对:</p>
<pre><code>" ".join([x for x in processing(s)])
</code></pre>