Lintcode：有效數字

問題：

給定一個字符串，驗證其是否爲數字。

樣例：

樣例 1:

輸入: "0"
輸出: true
解釋: "0" 可以被轉換成 0

樣例 2:

輸入: "0.1"
輸出: true
解釋: "0.1" 可以被轉換成 0.1

樣例 3:

輸入: "abc"
輸出: false

樣例 4:

輸入: "1 a"
輸出: false

樣例 5:

輸入: "2e10"
輸出: true
解釋: "2e10" 代表 20,000,000,000

python：

class Solution:
    """
    @param s: the string that represents a number
    @return: whether the string is a valid number
    """
    def isNumber(self, s):
        # write your code here
        if s == None:
            return False
        start = 0
        end = len(s)-1
        while start < end and s[start] == ' ':
            start += 1
        while end > start and s[end] == ' ':
            end -= 1
        if s[start] == '+' or s[end] == '-':
            start += 1
        if start > end:
            return False
        
        isNum = False
        isExp = False
        isDot = False
        while start <= end:
            if s[start].isdecimal():
                isNum = True
            elif s[start] == 'e':
                if isExp or not isNum:
                    return False
                isExp = True
                isNum = False
            elif s[start] == '.':
                if isDot or isExp:
                    return False
                isDot = True
            elif s[start] == '+' or s[start] == '-':
                if s[start-1] != 'e':
                    return False
            else:
                return False
            start += 1
        return isNum

C++：

class Solution {
public:
    /**
     * @param s: the string that represents a number
     * @return: whether the string is a valid number
     */
    bool isNumber(string &s) {
        // write your code here
        if(s == "")
        {
            return false;
        }
        int start = 0;
        int end = s.size()-1;
        while(start < end && s[start] == ' ')
        {
            start++;
        }
        while(end > start && s[end] == ' ')
        {
            end--;
        }
        
        if(s[start] == '+' || s[start] == '-')
        {
            start++;
        }
        if (start > end)
        {
            return false;
        }
        
        bool isNum = false;
        bool isExp = false;
        bool isDot = false;
        while(start <= end)
        {
            if(int(s[start] - '0') >= 0 && int(s[start] - '9') <= 0)
            {
                isNum = true;
            }else if(int(s[start] - 'e') == 0)
            {
                if(isExp || !isNum)
                {
                    return false;
                }
                isExp = true;
                isNum = false;
            }else if(int(s[start] - '.') == 0)
            {
                if(isDot || isExp)
                {
                    return false;
                }
                isDot = true;
            }else if(int(s[start] - '+') == 0 || int(s[start] - '-') == 0)
            {
                if(int(s[start-1] - 'e') != 0)
                {
                    return false;
                }
            }else{
                return false;
            }
            start++;
        }
 
        return isNum;
    }
};

PS：規則不太好理解。。。

PS：前後去空，正負檢測，2個e直接返回，點不能出現在e所有後面，加減號只能緊接着e後出現