如何计算字符串中连续重复的字符数

Question

Basically, what I want to achieve is to not allow passwords with lots of consecutively repeated chars/digits. I'm trying to write a go function to which I pass a string and a maximum allowed number of consecutively repeated characters, and it should tell me if it's surpassed or not.

I used to do it easily in javascript using the a regular expression in the following way:

var regexString = '(.)\\1{' + (maxAllowedRepetitions) + ',}';
var regex = new RegExp(regexString);
return regex.test(string)

... where maxAllowedRepetitions is the max limit. If the limit is 3 and a string is 'blablabla', it returns false. If it is blablaaaabla, it returns true since the character 'a' is repeated more than 3 times.

I'm finding it difficult to achieve the same thing using go regexp.
I don't mind not using regexp. I just need a good way to achieve this.
Any suggestions?

Answer 1

https://play.golang.org/p/HeK3f4uEvz

package main

import "fmt"

func main() {

    // invalid password with repetition
    password := "blablaaaabla"
    fmt.Printf("%s invalid ? %+v
", password, invalid(password))

    // correct password with no repetition
    password = "blabla"
    fmt.Printf("%s invalid ? %+v
", password, invalid(password))

    // valid password with unicode character
    password = "bla∞∞bla"
    fmt.Printf("%s invalid ? %+v
", password, invalid(password))

    // invalid password with unicode character
    password = "bla∞∞∞bla"
    fmt.Printf("%s invalid ? %+v
", password, invalid(password))
}

func invalid(s string) bool {
    var lastChar rune
    var lastCharCount = 0
    for _, c := range s {
        if c == lastChar {
            lastCharCount++
            if lastCharCount >= 3 {
                return true
            }
        } else {
            lastChar = c
            lastCharCount = 1
        }
    }

    return false
}

results

blablaaaabla invalid ? true
blabla invalid ? false
bla∞∞bla invalid ? false
bla∞∞∞bla invalid ? true