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包含至少两个 0 和至少一个 1 的所有二进制字符串的正则表达式？

发布于 2024-12-10 21:32:06 字数 67 浏览 6 评论 0原文

我想它会是（E0 * 0 * EUE1 * E）？其中 E 是我的字母表的集合，至少有 2 个 0 和至少 1 个 1

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始于初秋 2024-12-17 21:32:07

尝试这个表达式：

^(.*0.*0.*1.*)|(.*0.*1.*0.*)|(.*1.*0.*0.*)$

编辑
可以简化为：

^.*(0.*0.*1)|(0.*1.*0)|(1.*0.*0).*$

Try this expression:

^(.*0.*0.*1.*)|(.*0.*1.*0.*)|(.*1.*0.*0.*)$

EDIT
Could be simplified to:

^.*(0.*0.*1)|(0.*1.*0)|(1.*0.*0).*$

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满身野味 2024-12-17 21:32:07

如果您被允许使用前瞻，我将这样做：（在带有注释的 PHP 自由间距模式下。）

$re = '/
    # Binary strings that include at least two 0s and at least one 1.
    ^                  # Anchor to start of string.
    (?=(?:[^0]*0){2})  # at least two 0s. 
    (?=[^1]*1)         # at least one 1.
    [+\-]?             # Optional leading sign.
    [01]+              # Match string of binary digits.
    $                  # Anchor to end of string.
    /x';

请注意，您可以在字符串的开头放置任意数量的前瞻（其在逻辑 < 中工作） code>AND 方式），指定多个逻辑要求。

If you are allowed to use lookahead, here is how I'd do it: (in PHP free-spacing mode with comments.)

$re = '/
    # Binary strings that include at least two 0s and at least one 1.
    ^                  # Anchor to start of string.
    (?=(?:[^0]*0){2})  # at least two 0s. 
    (?=[^1]*1)         # at least one 1.
    [+\-]?             # Optional leading sign.
    [01]+              # Match string of binary digits.
    $                  # Anchor to end of string.
    /x';

Note that you may place any number of lookaheads at the beginning of the string (which work in a logical AND manner), to specify multiple logical requirements.

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