正则表达式用于增加/减少数字模式

Question

How can I test for an increasing or decreasing number pattern in a string? Assuming the string is just a number.

Ex:

var regexp = new RegExp(...) // something here
regexp.test("123456789") //true
regexp.test("121231234") //false
regexp.test("987654321") //true
regexp.test("5678901234") //true

The test would be for a cycle of digits really. How could this be achieved?

Answer 1

This mess of a regex is the best I can produce. It finds groups of ascending or descending numbers. You'd need to throw a starting and ending check on either end but, I didn't know what kind you want. For example you may want to have start line(^) and end line($) wrapped around this regex. It uses lookaheads which you may want to lookup if you are unfamiliar http://www.regular-expressions.info/lookaround.html .

(((?=01|12|23|34|45|56|67|78|89|90)\d)+|((?=
09|98|87|76|65|4|43|32|21|10)\d)+)\d

Answer 2

I'm not sure that you can achieve that with regex, but definitely can do that with reduce .

 const test = ["123456789", "121231234", "987654321", "5678901234"]; const results = test .map((item) => item.split('')) .map((item) => { let lastDigit = item[0]; return item.slice(1).reduce((flag, digit) => { const ret = flag && digit > lastDigit; lastDigit = digit; return ret }, true) }); console.log(results); 

Answer 3

If you actually need to use regexp to do this, then this is simplest solution that comes to my mind:

/(^0*1*2*3*4*5*6*7*8*9*$)|(^9*8*7*6*5*4*3*2*1*0*$)/

* means "any number of".