[leetcode] 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.

//模拟数字组成即可

 1 class Solution
 2 {
 3 public:
 4   bool isNumber(const char *s)
 5   {
 6     bool has_front_num = false, has_back_num = false, has_e = false;
 7     int len = strlen(s);
 8     int i = 0;
 9 
10     while(i < len && ' ' == s[i])
11       i++;
12 
13     if(i == len) return false;
14 
15     if(i < len && (s[i] == '+' || s[i] == '-'))
16       i++;
17 
18     while(i < len && isdigit(s[i]))
19     {
20       i++;
21       has_front_num = true;
22     }
23 
24     if(i < len && s[i] == '.')
25       i++;
26    
27     while(i < len && isdigit(s[i]))
28     {
29       i++;
30       has_front_num = true;
31     }
32 
33     if(i < len && (s[i] == 'e' || s[i] == 'E') && has_front_num)
34     {
35       i++;
36       has_e = true;
37       if(i == len) return false;
38     }
39 
40    
41     if(i < len && (s[i] == '+' || s[i] == '-') && has_e)
42       i++;
43    
44     while(i < len && isdigit(s[i]) && has_e)
45     {
46       i++;
47       has_back_num = true;
48     }
49 
50     while(i < len && s[i] == ' ')
51       i++;
52 
53     if(i == len && has_e && has_back_num)
54       return true;
55     else if(i == len && !has_back_num && !has_e && has_front_num)
56       return true;
57 
58     return false;
59   }
60 };