65 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.

class Solution {
  public:
    string parse(string s) {
      int n = s.size(), i = 0, j, k;
      while ( i < n ) {
        if ( s[i] != ' ' ) break;
        i++;
      }
      j = i;
      while ( j < n ) {
        if ( s[j] == ' ') break;
        j++;
      }
      k = j;
      while ( k < n ) {
        if ( s[k] != ' ' ) break;
        k++;
      }
      if ( k != n ) return "";
      return s.substr(i, j-i);
    }
    string abstract(string s) {
      int n = s.size(), i = 0;
      string res = "";
      while ( i < n ) {
        if ( s[i] - '0' < 0 or s[i] - '0' > 9 ) res += s[i];
        if ( s[i] - '0' >= 0 and s[i] - '0' <= 9 ) {
          if ( res.size() == 0 or res[res.size()-1] != '0' ) res += '0';
        }
        i += 1;
      }
      return res;
    }
    bool isNumber(string s) {
      // integer(positive, negative), decimal (0.5, .5, and 5. both counts), 1e-5, 1e05
      // prefix spaces and trailing spaces are allowed
      s = parse(s);
      if ( s == "" ) return false;
      string res = abstract(s);
      string part1[3] = {"", "+", "-"};
      string part2[4] = {"0", "0.0", ".0", "0."};
      string part3[4] = {"", "e0", "e+0", "e-0"};
      for ( int i1 = 0; i1 < 3; i1++ ) {
        for ( int i2 = 0; i2 < 4; i2++ ) {
          for ( int i3 = 0; i3 < 4; i3++ ) {
            if ( res == part1[i1] + part2[i2] + part3[i3] ) return true;
          }
        }
      }
      return false;
    }
};

Notes

1. Parsing the string by removing spaces. If it splits into more than 1 part, it is not a valid number.
2. Convert the string to an abstract form, using '0' to represent digit(s), and keep all the non-digit character.
3. The valid abstract form can be enumerated. Note the trick of using part1, part2 and part3 for comparison.