LeetCode 65 - Valid Number

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.

My Code

class Solution {
public:
bool isRealNumber(string s)
{
int len = s.length();
if (len == 0)
return false;
if (s == "-" || s == "." || s == "+" || s == "-." || s == "+.")
return false;

int i = 0;
if (s[i] == '-' || s[i] == '+')
i++;

int dotCnt = 0;
for (; i < len; i++)
{
if (s[i] == '.')
{
dotCnt++;
if (dotCnt >= 2)
return false;
}
else if (s[i] >= '0' && s[i] <= '9')
continue;
else
return false;
}

return true;
}

bool isInteger(string s)
{
int len = s.length();
if (len == 0)
return false;
if (s == "-" || s == "+")
return false;

int i = 0;
if (s[i] == '-' || s[i] == '+')
i++;

int dotCnt = 0;
for (; i < len; i++)
{
if (s[i] >= '0' && s[i] <= '9')
continue;
else
return false;
}

return true;
}

bool isNumber(string s) {
int len = s.length();
int start = 0;
while (s[start] == ' ' || s[start] == '\t')
start++;
int end = len - 1;
while (s[end] == ' ' || s[end] == '\t')
end--;
s = s.substr(start, end-start+1);

int pos = s.find('e');
if (pos == -1)
{
return isRealNumber(s);
}
else
{
return isRealNumber(s.substr(0, pos)) && isInteger(s.substr(pos + 1));
}
}
};

Runtime: 12 ms

Testcases

Input	Result
-1e-10	true
-	false
.	false
.01	true
-.01	true
1.	true
-1.	true
01	true
-01	true
1e	false
e1	false
e	false
1e0	true