65. Valid Number

description:

看给定的数字是否是合法数字
Note:

Example:

answer:

class Solution {
public:
    bool isNumber(string s) {
        int len = s.size();
        int left = 0, right = len - 1;
        bool eExisted = false;
        bool dotExisted = false;
        bool digitExisted = false;
        while (s[left] == ' ') ++left;
        while (s[right] == ' ') --right;
        if (left >= right && (s[left] < '0' || s[left] > '9')) return false;
        if (s[left] == '.') dotExisted = true;
        else if (s[left] >= '0' && s[left] <= '9') digitExisted = true;
        else if (s[left] != '+' && s[left] != '-') return false;
        for (int i = left + 1; i <= right - 1; ++i) {
            if (s[i] >= '0' && s[i] <= '9') digitExisted = true;
            else if (s[i] == 'e' || s[i] == 'E') {
                if (!eExisted && s[i - 1] != '+' && s[i - 1] != '-' && digitExisted) {
                    eExisted = true;
                }
                else return false;
            } else if (s[i] == '+' || s[i] == '-') {
                if (s[i - 1] != 'e' && s[i - 1] != 'E') return false;
            } else if (s[i] == '.') {
                if (!dotExisted && !eExisted) dotExisted = true;
                else return false;
            } else return false;
        }
        if (s[right] >= '0' && s[right] <= '9') return true;
        else if (s[right] == '.' && !dotExisted && !eExisted && digitExisted) return true;
        else return false;
    }
};

relative point get√：

hint :

情况很复杂
https://www.cnblogs.com/grandyang/p/4084408.html