擅长:python、mysql、java
<p>好吧,也许现在我明白你在找什么了:</p>
<pre><code>def replaceit(st, remove, put, pos):
outs = ""
count = 0
for letter in st:
if letter == remove:
count += 1
if count == pos:
outs += put
else:
outs += letter
else:
outs += letter
return outs
</code></pre>
<p>输出:</p>
^{pr2}$
<p>当然,您可以检查参数no2和no3是否是len()为1的字符串。在</p>