65. Valid Number

description:

65

Valid Number

12.7%

Hard

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.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.

my solution:

class Solution {
public:
	bool isNumber(string s) {
		int flagE = 0;
		int posE = 0;
		int posPoint = 0;
		int flagPoint = 0;
		int flagSpace = 0;
		int flagFront = 0;
		int flagBack = 0;
		int i = 0;
		while (isspace(s[i])) i++;
		if (i == s.length()) return false;
		if (s[i] == '+' || s[i] == '-') i++;  // check for sign
		for (; i != s.length(); i++) {
			if (s[i] == '.') {
				flagPoint++;
				posPoint = i;
				if (flagPoint > 1) return false;
			}
			else if (isspace(s[i])) {
				flagSpace++;
				break;
			}
			else if (!isdigit(s[i]) && !isspace(s[i])) {
				if (s[i] == 'e' || s[i] == 'E') {
					posE = i;
					flagE++;
					if (flagE > 1) return false;
					if (s[i + 1] == '+' || s[i + 1] == '-') i++; //check for sign
				}
				else return false;
			}
		}
		if (flagSpace != 0) {
			for (; i != s.length(); i++) {
				if (!isspace(s[i])) return false;
			}
		}
		if (flagE == 1 && flagPoint == 1) {
			if (posE < posPoint) return false;
		}
		if (flagE == 1) {
			i = posE - 1 > -1 ? posE - 1 : 0;
			while (!isdigit(s[i])) {
				i--;
				if (i == -1) return false;
			}
			i = posE + 1 < s.length() ? posE + 1 : s.length() - 1;
			while (!isdigit(s[i])) {
				i++;
				if (i == s.length()) return false;
			}
		}	
		if (flagPoint == 1) {
			i = posPoint - 1 > -1 ? posPoint - 1 : 0;
			while (!isdigit(s[i])) {
				i--;
				if (i == -1) flagFront = 1;
			}
			i = posPoint + 1 < s.length() ? posPoint + 1 : s.length() - 1;
			while (!isdigit(s[i])) {
				i++;
				if (i == s.length()) flagBack = 1;
			}
			if (flagBack == 1 && flagFront == 1) return false;
		}
		return true;
	}
};

better ways:

by yangf0722

class Solution {
public:
    bool isNumber(string str) {
        int state=0, flag=0; // flag to judge the special case "."
        while(str[0]==' ')  str.erase(0,1);//delete the  prefix whitespace 
        while(str[str.length()-1]==' ') str.erase(str.length()-1, 1);//delete the suffix whitespace
        for(int i=0; i

thought:

提交了12次终于过了，这道题其实算法并不难，但是对细心程度的要求比较高，需要考虑到每一种情况。

全而不漏感觉需要从最基础的开始考虑，比如从{[0-9]|.|*}+开始考虑，并且还要熟悉科学计数法的定义。