LeetCode 65：有效数字（C ++）

Question

我想在input_string语句中将regex_match传递到if之前对其进行左右修剪，但我没有尝试过boost trim_right and trim_left，但不确定如何解决。

我如何简单地修剪输入，以使前两个输入trim_right和trim_left成为有效数字？

"  0  "

当前输出

"  0.1  "

期望的输出

#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;
}

Answer 1

最大的问题是，

[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.

需要一个持久的字符串对象来引用结果。如果没有，那么在检查结果之后，将访问不存在的字符串，这是未定义的行为。这意味着您需要修改循环以使用

[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.

中的std::regex_match，例如

std::regex_match

这将使您花费一个副本，但这是不修改源数据而执行此操作的唯一方法。如果可以修改源数据，则可以使用

trim_copy

代替。