LeetCode [65] Valid Number

Validate if a given string can be interpreted as a decimal number.

Some examples:
"0" => true
" 0.1 " => true
"abc" => false
"1 a" => false
"2e10" => true
" -90e3   " => true
" 1e" => false
"e3" => false
" 6e-1" => true
" 99e2.5 " => false
"53.5e93" => true
" --6 " => false
"-+3" => false
"95a54e53" => false

Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. However, here is a list of characters that can be in a valid decimal number:

• Numbers 0-9
• Exponent - "e"
• Positive/negative sign - "+"/"-"
• Decimal point - "."

Of course, the context of these characters also matters in the input.

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.

• method1: a number can be  "-----(+/-)dddd.dddde(+/-)dddd-----"  ('-' denotes space, 'd' denotes digit)
• those characters must appear in order, if they exist
• if 'e' exists, there must exist digits before and after 'e'
• there must exist at least one digit
• there cannot exist other characters

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class Solution { public: bool isNumber(string s) { bool hasdot = false, hasdigitbeforedot = false, hasdigitafterdot = false; bool hase = false, hasdigitaftere = false; int n = s.size(), i = 0; //beginning space while(i<n && s[i]==' ') i++; //sign of number if(i<n && (s[i]=='+'||s[i]=='-')) i++; //digit before dot while(i<n && isdigit(s[i])){i++; hasdigitbeforedot = true;} //dot if(i<n && s[i]=='.'){ i++; hasdot = true; //digit after dot while(i<n && isdigit(s[i])){i++; hasdigitafterdot = true;} } //exp if(i<n && s[i]=='e'){ hase = true; i++; //sign of exp if(i<n && (s[i]=='+'||s[i]=='-')) i++; //digit after e while(i<n && isdigit(s[i])){i++; hasdigitaftere = true;} if(!hasdigitaftere) return false; } //ending space while(i<n && s[i]==' ') i++; return (i==n && (hasdigitbeforedot||hasdigitafterdot)); } };