LeetCode-Valid Number

Validate if a given string is numeric.

Some examples:
"0" => true
" 0.1 " => true
"abc" => false
"1 a" => false
"2e10" => true

Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one.

Solution:

Code:

<span style="font-size:14px;">class Solution {public:    bool isDigit(const char &c) {        int temp = c - '0';        if ((0 <= temp) && (temp <= 9))            return true;        return false;    }        bool isWithoutSign (const char *s, int &index) {        int countE = 0;        int countP = 0;        bool containBlank = false;        bool containDigit = false;        bool containSign = false;        bool beforeE = false;        bool afterE = false;        bool beforeP = false;        bool afterP = false;        for (int i = index; i < strlen(s); i++) {            if (s[i] == 'e') {                if (containBlank) return false;                countE++;                if (countE > 1) return false;            } else if (s[i] == '.') {                if (containBlank) return false;                if (countE) return false;                countP++;                if (countP > 1) return false;            } else if (isDigit(s[i])) {                if (containBlank) return false;                containDigit = true;                if (!countP) beforeP = true;                else afterP = true;                if (!countE) beforeE = true;                else afterE = true;            } else if ((s[i] == '+') || (s[i] == '-')) {                if (containBlank) return false;                if (containSign) return false;                if (s[i-1] != 'e') return false;                containSign = true;            } else if (s[i] == ' ') {                containBlank = true;            } else {                return false;            }        }        if ((countE) && (!(beforeE && afterE))) return false;        if ((countP) && (!(beforeP || afterP))) return false;        if (!containDigit) return false;        return true;    }        bool isNumber(const char *s) {        bool result = false;        int index = 0;        while ((s[index] == ' ')) {            index++;        }        if ((s[index] == '+') || (s[index] == '-')) {            index++;            result = isWithoutSign(s, index);        } else {            result = isWithoutSign(s, index);        }        return result;    }};</span>