如何用Java正则表达式匹配合法的算术运算序列？

Hey there! Let's break down why your current regex isn't working consistently for valid arithmetic inputs like 1 + 1 + 1, and fix it up to match exactly what you need for your project.

The Problem with Your Current Regex #

Your regex (\s*[0-9]+\s*[\+\-\*\/]+\s*[0-9]+\s*){1,} has two key issues:

• Incorrect grouping logic: The regex is designed to match chunks of [number][operator][number] and repeat that. But when you have a longer expression like 1 + 1 + 1, the first chunk matches 1 + 1, leaving + 1 leftover — which can't match the chunk pattern (since it starts with an operator, not a number). That's why 1 + 1 + 1 fails, while 1 + 12 + 1 might accidentally work due to unreliable regex engine backtracking.
• Unintended multiple operators: The [\+\-\*\/]+ part allows consecutive operators like 2 ++ 3 or 5 -- 2, which are invalid inputs for your calculator.

The Corrected Regex #

Here's a regex that will properly match valid arithmetic expressions (single numbers or chains like 2 + 2 - 3) and reject invalid ones like 2 + 2 -:

^\s*\d+\s*(?:[+*/-]\s*\d+\s*)*$

Let's break down what each part does:

• ^ and $: These anchor the regex to the start and end of the input string. This ensures we don't match partial inputs (like the first half of 2 + 2 -).
• \s*\d+\s*: Matches a number (\d+) with any amount of whitespace (\s*) before and after it — handling the starting number of your expression.
• (?:[+*/-]\s*\d+\s*)*: A non-capturing group (the ?: makes it non-capturing for efficiency) that matches:
  • A single valid operator ([+*/-] — we put - at the end of the character class to avoid it being interpreted as a range character)
  • Followed by whitespace, a number, and more whitespace
  • The * means this group can repeat 0 or more times — so it handles expressions with multiple operators, or even just a single number if needed.

Adjustments for Edge Cases #

• If you don't want to allow single-number inputs (e.g., you require at least one operator), change the * at the end to a +:

  ^\s*\d+\s*(?:[+*/-]\s*\d+\s*)+$
  

• If you need to support negative numbers (like -5 + 3), you'd need to tweak the regex to handle optional leading - on numbers, but that's an extra step if your project requires it.

Testing Your Examples #

Let's verify this regex against your test cases:

• 1 + 1: ✅ Matches successfully
• 1 + 1 + 1: ✅ Matches successfully (no more failure!)
• 1 + 12 + 1: ✅ Matches successfully
• 2 + 2 -: ❌ Fails to match (perfect, since it's invalid)

内容的提问来源于stack exchange，提问作者zvz