[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.

思考：需要考虑的情况很多。参考[这里]。

class Solution {
public:
    bool isNumber(const char *s) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        string num = s;
		int i=0;
		while(num[i]==' ')i++;
		if(num[i]=='-' || num[i]=='+')i++;
		int j=num.size()-1;
		while(num[j]==' ')j--;
		if(i<=j)
			num = num.substr(i,j-i+1);
		else return false;

		int dot = -1;
		int ee = -1;
		for(int i=0; i<num.size(); i++)
		{
			if(dot == -1 && num[i] == '.')
				dot = i;
			else if(ee == -1 && num[i] == 'e'){
				ee = i;
				if(num[i+1] == '-' || num[i+1]=='+')
					i++;
			}
			else{ 
				int tmpnum = num[i]-'0';
				if(0<=tmpnum && tmpnum<=9)continue;
				else return false;
			}
		}
		//xxx.xxexx
		string startstr,midstr,laststr;
		if(dot==-1 && ee==-1){//xxxx
			startstr = num;
			if(startstr.size()<1)return false;
		}else if(dot!=-1 && ee==-1){//xxx.xxx
			startstr = num.substr(0,dot);
			midstr=num.substr(dot+1);//.1,2.,0.1,0.0,2.0
			if(startstr.size()<1 && midstr.size()<1)
				return false;
		}else if(dot==-1 && ee!=-1){//xxxexxx
			startstr = num.substr(0,ee);
			if(startstr.size()<1)return false;
			if(num[ee+1] == '-' || num[ee+1]=='+')
				laststr = num.substr(ee+2);
			else
				laststr = num.substr(ee+1);
			if(laststr.size()<1)return false;
		}else{//xxx.xxexx
			if(dot>ee)return false;
			startstr = num.substr(0,dot);
			midstr=num.substr(dot+1,ee-dot-1);
			if(startstr.size()<1 && midstr.size()<1)
				return false;
			if(num[ee+1] == '-' || num[ee+1]=='+')
				laststr = num.substr(ee+2);
			else
				laststr = num.substr(ee+1);
			if(laststr.size()<1)return false;
		}
		return true;
    }
};