LeetCode OJ-- Valid Number **@

https://oj.leetcode.com/problems/valid-number/

判断给的串，是不是合理的 数字形式

主要问题在需求定义上吧

class Solution {
public:
    bool isNumber(const char *s) {
        if(s == NULL)
            return false;

        int index = ;
        // remove heading spaces
        while(s[index] != '\0' && s[index] == ' ')
            index++;

        // only has spaces
        if(s[index] == '\0')
            return false;

        // check + or - allowed
        if(s[index] == '+' || s[index] == '-')
            index++;

        // remove tailing spaces
        bool hasSpace = false;
        int tailIndex = ;
        for(int i = index; s[i] != '\0'; i++)
        {
            if(s[i] == ' ')
            {
                if(hasSpace == false)
                    tailIndex = i - ;
                hasSpace = true;
                continue;
            }
            else if(hasSpace && s[i] != ' ')
                return false;

            if(hasSpace == false)
                tailIndex = i;
        }

        // check only one . and e or digits allowed 
　　　　　// . e can't both exists. and 8.  is valid
　　　　　// before e and after e must has digits
　　　　　// + - before them must be e
        bool hasNum = false;
        bool hasDot = false;
        bool hasE = false;
        for(int i = index; i != tailIndex +  && s[i] != '\0'; i++)
        {
            if(s[i] >= '' && s[i] <= '')
                hasNum = true;

            else if(s[i] == '.')
            {
                if(hasDot || hasE)
                    return false;

                hasDot = true;
            }
            else if(s[i] == 'e')
            {
                if(hasE || hasNum == false)
                    return false;
                hasE = true;
                hasNum = false;
            }

            else if(s[i] == '+' || s[i] == '-')
            {
                if(!(i >  && s[i-] == 'e'))
                    return false;
                hasNum = false;
            }
            else
                return false;
        }

        return hasNum;
    }
};