[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;
}
};