I try to match/get all repetitions in a string. This is what I\'ve done so far:
var str = \'abcabc123123\';
var REPEATED_CHARS_REGEX = /(.).*\\1/gi;
console
The answer above returns more duplicates than there actually are. The second for loop causes the problem and is unnecessary. Try this:
function stringParse(string){
var arr = string.split("");
for(var i = 0; i<arr.length; i++){
var letterToCompare = arr[i];
var j= i+1;
if(letterToCompare === arr[j]){
console.log('duplicate found');
console.log(letterToCompare);
}
}
}
This solution may be used if you don't want to use regex:
function test() {
var stringToTest = 'find the first duplicate character in the string';
var a = stringToTest.split('');
for (var i=0; i<a.length; i++) {
var letterToCompare = a[i];
for (var j=i+1; j<a.length; j++) {
if (letterToCompare == a[j]) {
console.log('first Duplicate found');
console.log(letterToCompare);
return false;
}
}
}
}
test()
Well, I think falsetru had a good idea with a zero-width look-ahead.
'abcabc123123'.match(/(.+)(?=\1)/g)
// ["abc", "123"]
This allows it to match just the initial substring while ensuring at least 1 repetition follows.
For M42's follow-up example, it could be modified with a .*?
to allow for gaps between repetitions.
'abc123ab12'.match(/(.+)(?=.*?\1)/g)
// ["ab", "12"]
Then, to find where the repetition starts with multiple uses together, a quantifier ({n}
) can be added for the capture group:
'abcabc1234abc'.match(/(.+){2}(?=.*?\1)/g)
// ["abcabc"]
Or, to match just the initial with a number of repetitions following, add the quantifier within the look-ahead.
'abc123ab12ab'.match(/(.+)(?=(.*?\1){2})/g)
// ["ab"]
It can also match a minimum number of repetitions with a range quantifier without a max -- {2,}
'abcd1234ab12cd34bcd234'.match(/(.+)(?=(.*?\1){2,})/g)
// ["b", "cd", "2", "34"]