Leetcode---65. Valid number

Title Link: Valid number

Validate if a given string is numeric.

Some Examples:

Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one.

The requirement of this question is to determine whether the 1 string is a number.

string processing, but there are several details to note:

1. Numbers can have leading and trailing spaces, but spaces are not allowed in the middle of a number;
2. For '. ', the maximum allowed is 1 times, the front can have no number, but must have a number behind;
3. For ' E ', it is allowed to appear at most 1 times, and must have numbers before and after it, but it is an integer, which cannot appear '. ' ；
4. For ' + ' and '-', ' e ' is allowed to appear up to 1 times before and after, and must appear at the front of the number;
5. As for the other characters, only the number (0~9) is allowed.

Here are a few examples:

"e"     =>   false"."     =>   false" "     =>   false".1"    =>   true"0e"    =>   false"e9"    =>   false"7e+6"  =>   true"6+1"   =>   false

Time complexity: O (N)

Space complexity: O (1)

1 class Solution2 {3  Public:4     BOOL Isnumber(string s)5     {6         int I = 0;7         //Skip leading spaces8          for( ; I < s.size() && "' == s[I]; ++ I);9         //Process signTen         if(' + ' == s[I] || '-' == s[I]) One             ++ I; A         //Process The following number part -         BOOL Digit = false, Dot = false, Exp = false; -          for( ; I < s.size(); ++ I) the         { -             if('. ' == s[I] && !Dot) //'. ' Cannot appear 2 times, '. ' There can be no numbers in front -                 Dot = true; -             Else if(' E ' == s[I] && !Exp && Digit) //' E ' cannot appear 2 times, ' E ' must be preceded by a number +             { -                 //' E ' cannot appear after '. ', after ' e ' must be an integer (can be positive or negative) +                 Dot = Exp = true; A                 if(I + 1 < s.size() && (' + ' == s[I + 1] || '-' == s[I + 1])) at                     ++ I; -                 if(I + 1 >= s.size() || !(s[I + 1] >= ' 0 ' && s[I + 1] <= ' 9 ')) -                     return false; -             } -             Else if(s[I] >= ' 0 ' && s[I] <= ' 9 ') -                 Digit = true; in             Else -                  Break; to         } +          -         //Skip back spaces the          for( ; I < s.size() && "' == s[I]; ++ I); *          $         return Digit && I == s.size();Panax Notoginseng     } - };

Reprint please indicate source: Leetcode---65. Valid number

Leetcode---65. Valid number