Leetcode 65. 有效数字

难点在于考虑周全,尝试了很多次,根据错误案例提示写全了

class Solution {
public:
    bool isnum(string s) {
        if (s[0] == '-' || s[0] == '+') return s.size() != 1 &&
            ((s[1] >= '0' && s[1] <= '9') || s[1] == '.') && isnum(string(s, 1, 999));
        for (auto &x : s)
            if (x<'0' || x>'9') return false;
        return true;
    }
    bool isf(string s) {
        if (s.empty()) return false;
        if (s[0] == '-' || s[0] == '+') return s.size() != 1 &&
            ((s[1] >= '0' && s[1] <= '9') || s[1] == '.') && isf(string(s, 1, 999));
        int k = s.find('.');
        if (k == string::npos) return isnum(s);
        else if (k == s.size() - 1) return  s.size() != 1 && isnum(string(s, 0, k));
        else return  isnum(string(s, 0, k)) && s[k + 1] >= '0'&&s[k + 1] <= '9'&&isnum(string(s, k + 1, 999));
    }
    bool isNumber(string s) {
        int t = 0;
        while (t<s.size() && s[t] == ' ') ++t;
        if (t) return isNumber(string(s, t, 999));
        while (!s.empty() && s.back() == ' ')s.pop_back();
        int k = s.find_first_of("Ee");
        if (k == string::npos) return isf(s);
        else return k != s.size() - 1 && isf(string(s, 0, k)) && isnum(string(s, k + 1, 999));
    }
};

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转载自blog.csdn.net/bendaai/article/details/80266877
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