正则匹配顺序连续或者倒序连续的字符串及其相应的个数

Question

面试中遇到这样一道题，有一个字符串，"12345678abcABCDefghijk9874321YXWV321"，现在定义同类型字符（数字为一类型，大写字母为一类型，小写字母为一类型）4个及4个以上为连续，需要找出连续（包括顺序连续和倒序连续）的个数，请问一下各位大神这个应该怎么解决呢？谢谢。

Answer 1

连续的意思是上一个字符和下一个字符的ascii值相差1，并且连续出现4次和以上。这个貌似c语言作业...

java代码:

String str = "12345678abcABCDefghijk9874321YXWV321";

        char f=str.charAt(0);

        int count = 0;

        for(int i=1 ; i < str.length(); ++i) {

            char c = str.charAt(i);

            if(c-f == 1 || f-c==1) {

                ++count;

            }else {

                if(count >= 3) {

                    System.out.println(  str.substring(  i-count-1 ,i)  ) ;

                }

                count = 0;

            }

            f=c;

            

        }

Answer 2

function trans (str) {

  let before = ''

  let len = 0

  let order = null

  let matched = []

  for (let i = 0, length = str.length; i < length; i++) {

    let cur = str[i]

    if (len === 0) {

      before = cur

      len = 1

      order = null

      continue

    }

    let diff = cur.charCodeAt(0) - before.charCodeAt(0)

    if (Math.abs(diff) === 1) {

      order = order || diff

      if (order === diff) {

        len += 1

        before = cur

        continue

      }

    }

    if (len >= 4) {

      matched.push(str.slice(i - len, i))

    }

    before = cur

    len = 1

    order = null

  }

  if (len >= 4) {

    matched.push(str.slice(str.length - len))

  }

  return {

    count: matched.length,

    matched

  }

}

trans("12345678abcABCDefghijk9874321YXWV321").count