LeetCode 65：有效数字（C++）

Question

我想在if 语句中将input_string 向左/向右修剪，然后将其传递给regex_match，我尝试boost trim_right and trim_left 失败，但不知道如何解决它。

我如何简单地修剪输入，以便前两个输入 " 0 " 和 " 0.1 " 成为有效数字？

#include <string>
#include <boost/algorithm/string.hpp>

#include <iostream>
#include <regex>
#include <vector>
using namespace std;

int main() {
    vector<string> string_vector = {"   0  ","   0.1   ","abc","1 a","2e10","-90e3","1e","e3","6e-1","99e2.5","53.5e93","--6","-+3","95a54e53"};

    regex expression_two("^[+-]?(?:[0-9]*\\.[0-9]+|[0-9]+\\.[0-9]*|[0-9]+)[Ee][+-]?[0-9]+$|^[+-]?(?:[0-9]*\\.[0-9]+|[0-9]+\\.[0-9]*|[0-9]+)$|^[+-]?[0-9]+$");

    for (const auto &input_string: string_vector) {
        if (std::regex_match(input_string, expression_two))
            cout << "[0-9] Char Class: '" << input_string << "' is a valid number." << endl;
    }
    return 0;
}

电流输出

[0-9] Char Class: '2e10' is a valid number.
[0-9] Char Class: '-90e3' is a valid number.
[0-9] Char Class: '6e-1' is a valid number.
[0-9] Char Class: '53.5e93' is a valid number.

所需的输出

[0-9] Char Class: '  0  ' is a valid number.
[0-9] Char Class: '  0.1  ' is a valid number.
[0-9] Char Class: '2e10' is a valid number.
[0-9] Char Class: '-90e3' is a valid number.
[0-9] Char Class: '6e-1' is a valid number.
[0-9] Char Class: '53.5e93' is a valid number.

Answer 1

最大的问题是std::regex_match 需要一个持久的字符串对象来引用结果。如果您没有，那么在检查结果之后将访问一个不存在的字符串，这是未定义的行为。这意味着您需要修改循环以使用 trim_copy from What's the best way to trim std::string? like

for (const auto &input_string: string_vector) {
    auto trimmed = trim_copy(input_string);
    if (std::regex_match(trimmed, expression_two))
        cout << "[0-9] Char Class: '" << input_string << "' is a valid number." << endl;
}

这将花费您一份副本，但这是在不修改源数据的情况下执行此操作的唯一方法。如果您可以修改源数据，那么您可以使用

for (auto &input_string: string_vector) {
    if (std::regex_match(trim(input_string), expression_two))
        cout << "[0-9] Char Class: '" << input_string << "' is a valid number." << endl;
}

改为。