[LeetCode]Valid Number

正则表达式验证字符串是否为数字

最新推荐文章于 2019-04-07 16:41:47 发布

转载最新推荐文章于 2019-04-07 16:41:47 发布 · 82 阅读

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原文链接：http://www.cnblogs.com/Sean-le/p/4740457.html

本文介绍了一个使用状态机实现的正则表达式，用于验证给定字符串是否为数字。包括特殊符号和科学计数法的支持。

Valid Number

alidate 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.

参考大神的状态机写的。这种还是正面来比较好。

 1 class Solution
 2 {
 3 public:
 4     bool isNumber(string s)
 5     {
 6         enum InputType
 7         {
 8             INVALID,    // 0
 9             SPACE,      // 1
10             SIGN,       // 2
11             DIGIT,      // 3
12             DOT,        // 4
13             EXPONENT,   // 5
14             NUM_INPUTS  // 6
15         };
16         
17         int transitionTable[][NUM_INPUTS] =
18         {
19             -1,  0,  3,  1,  2, -1,     // next states for state 0
20             -1,  8, -1,  1,  4,  5,     // next states for state 1
21             -1, -1, -1,  4, -1, -1,     // next states for state 2
22             -1, -1, -1,  1,  2, -1,     // next states for state 3
23             -1,  8, -1,  4, -1,  5,     // next states for state 4
24             -1, -1,  6,  7, -1, -1,     // next states for state 5
25             -1, -1, -1,  7, -1, -1,     // next states for state 6
26             -1,  8, -1,  7, -1, -1,     // next states for state 7
27             -1,  8, -1, -1, -1, -1,     // next states for state 8
28         };
29         int i=0;
30         int state = 0;
31         while (i<s.length())
32         {
33             InputType inputType = INVALID;
34             if (isspace(s[i]))
35                 inputType = SPACE;
36             else if (s[i] == '+' || s[i] == '-')
37                 inputType = SIGN;
38             else if (isdigit(s[i]))
39                 inputType = DIGIT;
40             else if (s[i] == '.')
41                 inputType = DOT;
42             else if (s[i] == 'e' || s[i] == 'E')
43                 inputType = EXPONENT;
44             
45             state = transitionTable[state][inputType];
46             
47             if (state == -1)
48                 return false;
49             else
50                 i++;
51         }
52         
53         return state == 1 || state == 4 || state == 7 || state == 8;
54     }
55 };