所以“只有一次”部分是关键。列出所有排列显然是不可行的。如果您的语言/环境支持前瞻和反向引用，您可以让它更容易一些：
r(?=[aeip]{4,4})(.)(?!\1)(.)(?!\1|\2)(.)(?!\1|\2|\3).r
仍然很丑，但它是这样工作的：
r # match an r
(?= # positive lookahead (doesn't advance position of "cursor" in input string)
[aeip]{4,4}
) # make sure that there are the four desired character ahead
(.) # match any character and capture it in group 1
(?!\1)# make sure that the next character is NOT the same as the previous one
(.) # match any character and capture it in group 2
(?!\1|\2)
# make sure that the next character is neither the first nor the second
(.) # match any character and capture it in group 3
(?!\1|\2|\3)
# same thing again for all three characters
. # match another arbitrary character
r # match an r
Working demo.
这既不优雅也不可扩展。因此，您可能只想使用r([aiep]{4,4})r（捕获四个关键字母）并确保没有正则表达式的附加条件。
编辑：事实上，如果您只想确保有 4 个不相同的字符，上述模式才是真正有用和必要的。对于您的具体情况，再次使用前瞻，有更简单（尽管更长）的解决方案：
r(?=[^a]*a[^a]*r)(?=[^e]*e[^e]*r)(?=[^i]*i[^i]*r)(?=[^p]*p[^p]*r)[aeip]{4,4}r
解释：
r # match an r
(?= # lookahead: ensure that there is exactly one a until the next r
[^a]* # match an arbitrary amount of non-a characters
a # match one a
[^a]* # match an arbitrary amount of non-a characters
r # match the final r
) # end of lookahead
(?=[^e]*e[^e]*r) # ensure that there is exactly one e until the next r
(?=[^i]*i[^i]*r) # ensure that there is exactly one i until the next r
(?=[^p]*p[^p]*r) # ensure that there is exactly one p until the next r
[aeip]{4,4}r # actually match the rest to include it in the result
Working demo.
对于具有deee 池的r....m，可以将其调整为：
r(?=[^d]*d[^d]*m)(?=[^e]*(?:e[^e])*{3,3}m)[de]{4,4}m
这样可以确保只有一个 d 和 3 个 es。
Working demo.