JavaScript Regular Expression Rule for repeated characters

Question

I am trying to create a RegEx rule to find side-to-side digits in a number. For example given an array of:

const nums = [1, 2, 33, 4, 22, 5, 66, 112];

I want to remove the digits [33, 22, 66, 112] from the array because they have repeated digits.

I tried the /[0-9/{2} but this seems to not work.

Answer 1

You can use filter with regex pattern ( which uses capturing group and back reference )

• [0-9]{2} - Means match any digit from 0 to 9 two times which doesn't guaranty repeated digits

• ([0-9])\\1

  • ([0-9]) - Means match digit 0 to 9 ( captured group 1 )
  • \\1 - should match the same value as the captured group 1

 const nums = [1, 2, 33, 4, 22, 5, 66, 112]; let nonRepeated = nums.filter(num => !/([0-9])\\1/.test(""+num)) // can replace with !/([0-9])\\1/.test(num) because it implicit coerce to string console.log(nonRepeated)

Answer 2

像这样使用\\1反向引用：

/(\d)\1+/

Answer 3

You can use filter and includes

const nums = [1, 2, 33, 4, 22, 5, 66, 112];

result = nums.filter(v => ![33, 22, 66, 112].includes(v));

Answer 4

the shortest approach is just to check if a same character is followed by itselft. No need for checking dor digits, because the value contains only digits or a dot (or E or space). The last occurs only once.

 var nums = [1, 2, 33, 4, 22, 5, 66, 112], nonRepeated = nums.filter(v => !/(.)\\1/.test(v)); console.log(nonRepeated);