regex - Regular expression pattern to match string without any 2 consecutive repeated characters -

i can write regular expression match string contains 2 consecutive repeated characters:

/(\w)\1/ 

how do complement of that? want match strings don't have 2 consecutive repeated characters. i've tried variations of following without success:

/(\w)[^\1]/ ;doesn't work hoped /(?!(\w)\1)/ ;looks ahead, portion of string match /(\w)(?!\1)/ ;again, portion of string match 

i don't want language/platform specific way take negation of regular expression. want straightforward way this.

the below regex match strings don't have repeated characters.

^(?!.*(\w)\1).* 

(?!.*(\w)\1) negative lookahead asserts string going matched won't contain repeated characters. .*(\w)\1 match string has repeated characters @ middle or @ start or @ end. ^(?!.*(\w)\1) matches starting boundaries except 1 has repeated characters. , following .* matches characters exists on particular line. note this matches empty strings also. if don't want match empty lines change .* @ last .+

note ^(?!(\w)\1) checks repeated characters @ start of string or line.

lookahead , lookbehind, collectively called "lookaround", zero-length assertions start , end of line. not consume characters in string, assert whether match possible or not. lookaround allows create regular expressions impossible create without them, or longwinded without them.