亚洲国产日韩欧美一区二区三区,精品亚洲国产成人av在线,国产99视频精品免视看7,99国产精品久久久久久久成人热,欧美日韩亚洲国产综合乱

Consider extraction without occupancy and use look-around to extract subgroups: (?=.*?((12|21)2{4}))

Demo link: http://regex.zjmainstay.cn/r/...

>>> import re
>>> ss='1212121212121212121212'
>>> re.findall(r'(?=((12|21){4}))',ss)
[('1212121212', '12'), ('2121212121', '21'), ('1212121212', '12'), ('2121212121', '21'), ('1212121212', '12'), ('2121212121', '21'), ('1212121212', '12'), ('2121212121', '21'), ('1212121212', '12'), ('2121212121', '21'), ('1212121212', '12'), ('2121212121', '21'), ('1212121212', '12')]

js's Re engine is a bit rough and needs to loop back to submatches.

var str="1212121212121212121212";
var pattern=/(?=((12|21){4}))/g;
while(m = pattern.exec(str)){
    console.log(m[1])
    pattern.lastIndex++ //由于沒(méi)有消耗字符，js的Re引擎不會(huì)遞增索引。
}

The question is unclear, I can only guess. .

(([0-9]){2})+

/((12){5})|((21){5})/.test(str)

var pattern=/(?:(1)(?=(?:21){4}2))|(?:(2)(?=(?:12){4}1))/g;
var str="1212121212121212121212";
console.log(str.match(pattern));

/(w+)1{4}/