[Leetcode 27] 65 Valid Number

Problem:

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.

 

Analysis:

So many cases to consider!

First of all, remove all the space at both ends of the given string;

Now we start from the left. 

The first position is a little bit special

1. '+/-': skip them

2. '.': set the dotFlag

3. '0~9': set the numFlag

4. others:  invalid input 

 

we continue 

if '.' is met, we break out of the current loop, start a new one. Note after '.', only 'e', numbers are allowed.

if 'e' is met, break out of the current loop, start a new one. Note after 'e', only numbers or '+/-' are allowed

 

Code:

 1 class Solution {
 2 public:
 3     bool isNumber(const char *s) {
 4         // Start typing your C/C++ solution below
 5         // DO NOT write int main() function
 6         int idx = 0, len = strlen(s)-1;
 7         bool dotFlag = false, numFlag = false, eFlag=false;
 8         
 9         if (s=="") return false;
10         
11         while (s[idx] == ' ') idx++;
12         if (idx > len) return false;
13         
14         while (s[len] == ' ') len--;
15         
16         
17         if (s[idx] == '+' || s[idx] == '-') idx++;
18         
19         if (idx > len) return false;
20         
21         int i;
22         for (i=idx; i<=len; i++) {
23             if (s[i] == 'e') {
24                 i += 1;
25                 eFlag = true;
26                 break;
27             }
28             
29             if (s[i] == '.') {
30                 i += 1;
31                 dotFlag = true;
32                 break;
33             }
34             
35             if (s[i]>='0' && s[i]<='9') {
36                numFlag = true;
37             } else {
38                 return false;
39             }
40             
41         }
42         
43         
44         if (dotFlag) {
45             if (!numFlag && (i>len || ((s[i]<'0' && s[i]>'9') && s[i]!='e')))
46                 return false;
47             
48             for (; i<=len; ++i) {
49                 if (s[i] <'0' || s[i] > '9') {
50                     if (s[i] == 'e') {
51                         i+= 1;
52                         eFlag = true;
53                         break;
54                     } else {
55                         return false;
56                     }
57                 }
58                 
59                 numFlag = true;
60             }
61         }
62         
63         //number after e
64         if (eFlag) {
65             if (!numFlag) return false;
66             
67             if (i==idx+1 || i>len) return false;
68         
69             if (s[i] == '+' || s[i] == '-')
70                 i += 1;
71             
72             if (i > len) return false;
73             
74             for (; i<=len; i++) {
75                 if (s[i]<'0' || s[i]>'9')
76                     return false;
77             }
78         }
79         
80         
81         return true;
82     }
83 };
View Code

 

Attention:

posted on 2013-05-19 13:41  freeneng  阅读(186)  评论(0)  编辑  收藏  举报

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