interview encountered such a question, there is a string, "12345678abcABCDefghijk9874321YXWV321", now define the same type of characters (numbers for one type, uppercase letters for one type, lowercase letters for one type) for 4 or more consecutive, need to find out the number of consecutive (including sequential and reverse continuous), I would like to ask you how to solve this? Thank you.
continuous means that the ascii value of the previous character is 1 different from that of the next character, and it appears 4 or more times in a row. This looks like a c language assignment.
java Code:
String str = "12345678abcABCDefghijk9874321YXWV321";
char f=str.charAt(0);
int count = 0;
for(int i=1 ; i < str.length(); PPi) {
char c = str.charAt(i);
if(c-f == 1 || f-c==1) {
PPcount;
}else {
if(count >= 3) {
System.out.println( str.substring( i-count-1 ,i) ) ;
}
count = 0;
}
f=c;
}function trans (str) {
let before = ''
let len = 0
let order = null
let matched = []
for (let i = 0, length = str.length; i < length; iPP) {
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 happens to be learning rules recently. I think it's appropriate to use regular matching
let str = "12345678abcABCDefghijk9874321YXWV321"
let aa = str.match (/ d {4,} / g)
let bb = str.match (/ [Amurz] {4,} / g)
let cc = str.match (/ [arelz] {4,} / g)
returns arrays that match numbers, uppercase and lowercase, respectively.
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