42 wins the offer series: a string value

The big question is not difficult, the main consideration several aspects:

1. The sign bit, can occur only once, only appear in front of the digital

2. decimal point, can appear only once, it must appear at the end of numbers

3. After the index e or E, can occur before the digital or digital

4. The necessary condition is called Digital must contain a number (e count does not count and E may be set according to the subject, this time is not assumed)

. 1  class Solution {
 2  public :
 . 3      BOOL that isNumeric ( char * STR)
 . 4      {
 . 5          IF (STR == NULL || STR [ 0 ] == ' \ 0 ' )
 . 6              return  to false ;
 . 7          int SIG = 0 ; // Analyzing the number of symbol bits 
. 8          int NUM = 0 ; // number of digits 
. 9          int Point = 0 ; // number of decimal
 10  
. 11          @Analyzing the first space, skip space 
12 is          the while (* STR == '  ' )
 13 is              STR ++ ;
 14  
15          // determines the symbol bit 
16          the while (* STR == ' + ' || * STR == ' - ' )
 . 17          {
 18 is              ++ SIG ;
 . 19              STR ++ ;
 20 is              IF (SIG> . 1 )
 21 is              {
 22 is                  return  to false ;
 23 is              }
 24          }
25          // number of digits is determined 
26 is  
27          the while (* STR> = ' 0 ' && * STR <= ' . 9 ' )
 28          {
 29              NUM ++ ;
 30              STR ++ ;
 31 is          }
 32          the while (* STR == ' . ' ) / / what I wrote, that strange results can not pass the test case, see below
 33                             // the while (NUM && * str == '.')
 34                             // in front of the decimal point must have digital is digital it's not a very simple truth it 
35          {
 36              Point ++ ;
37             str++;
38             if (sig > 1)
39             {
40                 return false;
41             }
42         }
43         while (*str >= '0' && *str <= '9')
44         {
45             num++;
46             str++;
47         }
48         if (*str == 'e' || *str == 'E' )
 49          {
 50              str ++ ;
51              {
 52                  themselves = 0 ;
53                  while (* str == ' + ' || * str == ' - ' )
 54                  {
 55                      itself ++ ;
56                      str ++ ;
57                      IF (at> 1 )
 58                      {
 59                          return  false ;
60                     }
 61                  }
 62                  // number determination number 
63 is                  NUM = 0 ;
 64                  the while (* STR> = ' 0 ' && * STR <= ' . 9 ' )
 65                  {
 66                      NUM ++ ;
 67                      STR ++ ;
 68                  }
 69                  IF (NUM = = 0 ) // rear e must digital 
70                      return  to false ;
 71 is                  IF (STR == *' \ 0 ' )
 72                      return  to true ;
 73 is                  the else 
74                  {
 75                      return  to false ;
 76                  }
 77              }
 78          }
 79          the else 
80              IF (STR * == ' \ 0 ' && NUM) // must include a digital number is
 81                  return  to true ;
 82              the else 
83              {
 84                  return  to false ;
 85             }
86         
87     }
88 
89 };

 

There is a test case, "-.123" sample output turned out to be true?

Is that right figures, cattle passenger network in the end how you think, not in front of the decimal point can it?