至少有一个 1 和偶数个 0 的二进制字符串的正则表达式

Question

I want to find a regex binary string expression that has at least one 1 and an even number of 0.

For strings that has an even number of 0, I have:

1*(01*01*)*

For strings with at least one 1, I have:

0*1(0+1)*

However, I struggle to combine the two together. Can anyone give me a hint on how to do it? thanks!

Answer 1

Without any fancy regex stuff, just ^ start, $ end anchors and | pipe , how about

^(00)*(1|01+0)(1*01*0)*1*$

Should be self explanatory. Play with it at regex101 . Hope I've not overlooked anything:)

If lookarounds are allowed, I could also think of a negative one: ^(??0*$)(::1*01*0)*1*$

Answer 2

One option to match an even number of zeroes and at least a single 1 might be using a positive lookahead (?= (if supported)

^(?=(?:1*01*01*)+$)0*1[10]*$

• ^ Start of string
• (?= Positive lookahead, assert what is on the right is
  • (?:1*01*01*)+$ Repeat 1+ times matching two times a zero with optional one's
• ) Close lookahead
• 0*1 Match zere or more times a zero, then match the required one
• [10]* Match 0+ times a zero or one
• $ End of string

Regex demo

---

To also match only one's (so without a zero) you could use

^(?=(?:(?:1*01*01*)+|1+)$)0*1[10]*$

Regex demo

Answer 3

I'm not sure if I understand your requirements correctly, but if + should stand for alternation (normally | ), and only * , + , and brackets are allowed (as your comment below suggests) then

(1*01*01*)*(11*+011*0)(1*01*01*)*

should work.

The idea is that there must exist a 1 such that either left and right from it must be an equal number of 0 s or left and right from it must be an odd number of 0 s: (11*+011*0)

Your 0*1(0+1)* would also match 01 so I believe it doesn't work.

In either case left and right from such a group there can exist an equal number of 0 s with an arbitrary number 1 s in between: (1*01*01*)* .

In most regex implementations the above regex would be written as

^(1*01*01*)*(11*|011*0)(1*01*01*)*$ .