php - Regular match 5 consecutive 1212121212 or 2121212121, numbers can be reused

Question

For example string 1212121212121212121212

It needs to match 5 consecutive 12s or 5 consecutive 21s to be considered qualified

The number of matches should be 13, as shown in the figure below, the red lines represent matches.

Answer 1

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

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

Answer 2

>>> 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++ //由于没有消耗字符，js的Re引擎不会递增索引。
}

Answer 3

The question is unclear, I can only guess. .

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

Answer 4

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

Answer 5

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

Answer 6

/(w+)1{4}/