Something like this would work:
"aaabbcapppp".match(/(.)1*/g) // ["aaa", "bb", "c", "a", "pppp"]
The (.)
matches any single character, captured in group 1, followed by that same character repeated zero or more times (1
is a backreference which matches exactly what was matched in group 1).
To match only Latin letters, consider using [a-z]
, for example:
"aaa-bbca!!pppp".match(/([a-z])1*/g) // ["aaa", "bb", "c", "a", "pppp"]
Here, the -
and !!
are not included in the result array.
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