题解 | #牛的回文编号II#

#include <cstddef>
#include <iterator>
#include <string>
class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param x string字符串
     * @return bool布尔型
     */
    bool isPalindromeNumber(string x) {
        std::size_t dotPos = x.find('.');

        std::string integerPart = x.substr(0, dotPos);
        std::string decimalPart = (dotPos != std::string::npos) ? x.substr(
                                      dotPos + 1) : "";

        decimalPart.erase(std::find_if(std::rbegin(decimalPart),
        std::rend(decimalPart), [](const auto & ch) -> bool{
            return ch != '0';
        }).base(), std::end(decimalPart));

        bool integerIsPalindrome = isPalindrome(integerPart);
        bool decimalIsPalindrome = decimalPart.empty() ? true : isPalindrome(
                                       decimalPart);

        return integerIsPalindrome && decimalIsPalindrome;
    }

    bool isPalindrome(std::string str) {
        int left{};
        int right = str.size() - 1;

        for (; left <= right; left++, right--) {
            if (str[right] != str[left]) {
                return false;
            }
        }

        return true;
    }
};