[leetcode]Valid Number

看着非常简单,leetcode上面通过率极低...

可以用正则表达式，不过面试肯定不是让你写个就ok了。。

所以我们自己来构造DFA吧。。。

写个表TAT

在各种数据的调试下，更改状态才过了（详情见照片上的修改

enum Op{
    Operator, //0
    Digit, // 1
    E, // 2
    Dot, //3
    Space,// 4
    Invalid
};
int table[][5] = {
    {1,2,-1,8,-1},
    {-1,2,-1,8,-1},
    {-1,2,4,3,-1},
    {-1,7,4,-1,-1},
    {5,6,-1,-1,-1},
    {-1,6,-1,-1,-1},
    {-1,6,-1,-1,-1},
    {-1,7,4,-1,-1},
    {-1,9,-1,-1,-1},
    {-1,9,4,-1,-1}

};
class Solution {
public:
    bool trim(string& s) {
        int start = 0;
        int end = s.size() - 1;
        while(start <= end && s[start] == ' ') start++;
        while(end >= 0 && s[end] == ' ') end--;
        if(end >= start) s = s.substr(start , end - start + 1);
        else return false;
        return true;
    }
    Op getOp(char ch) {
        if(ch == '+' || ch == '-') return Operator;
        if(ch >= '0' && ch <= '9') return Digit;
        if(ch == 'e') return E;
        if(ch == '.') return Dot;
        if(ch == ' ') return Space;
        return Invalid;
    }
    bool isNumber(const char *s) {
        string str = s;
        if(!trim(str)) return false;
        int pos = 0;
        int size = str.size();
        int state = 0;
        while(pos < size) {
            char ch = str[pos];
            Op x = getOp(ch);
            if(x == Invalid) return false;
            state = table[state][(int)x];
            if(state == -1) return false;
            pos++;
        }
        return state == 2 || state == 6  || state == 7 || state == 3 || state == 9;
    }
};