[Leetcode] Valid Number

Question:

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.

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Solution 1:

利用现有库，String.replaceAll(),关于的正则表达式问题可以参看前一篇博客。

 1 public class Solution {
 2     public boolean isNumber(String s) {
 3         s=s.trim();
 4         s=s.replaceAll("e|E", "/");
 5         s=s.replaceAll("[a-zA-Z]", "x");
 6         s=s.replaceAll("/", "e");
 7     
 8         try {
 9             Double.parseDouble(s);
10         } catch (Exception e) {
11             // TODO: handle exception
12         
13             return false;
14         }
15     
16         return true;
17     }
18 }
19

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Solution2:

Finite-state Machine solution（有限状态机）。

代码转自：https://github.com/AnnieKim/LeetCode/blob/master/ValidNumber.h

 1 class Solution {
 2 public:
 3     bool isNumber(const char *s) {
 4         enum InputType {INVALID, SPACE, SIGN, DIGIT, DOT, EXPONENT};
 5         int transitionTable[][SPACEEND] = 
 6         { /* 0   1   2   3   4   5  */
 7              0,  1,  2,  3,  4,  0, // 0: INVALID
 8              0,  1,  2,  3,  4,  0, // 1: SPACE
 9              0,  0,  0,  3,  4,  0, // 2: SIGN
10              0,  6,  0,  3,  7,  5, // 3: DIGIT
11              0,  0,  0,  7,  0,  0, // 4: DOT
12              0,  0,  2,  8,  0,  0, // 5: EXPONENT
13              0,  6,  0,  0,  0,  0, // 6: END WITH SPACE
14              0,  6,  0,  7,  0,  5, // 7: DOT AND DIGIT
15              0,  6,  0,  8,  0,  0, // 8: END WITH SPACE OR DIGIT
16         };
17         
18         InputType last = INVALID;
19         while (*s != '')
20         {
21             InputType state = INVALID;
22             if (*s == ' ')
23                 state = SPACE;
24             else if (isdigit(*s))
25                 state = DIGIT;
26             else if (*s == '+' || *s == '-')
27                 state = SIGN;
28             else if (*s == 'e')
29                 state = EXPONENT;
30             else if (*s == '.')
31                 state = DOT;
32             last = (InputType) transitionTable[last][state];
33             if (last == INVALID) return false;
34             s++;
35         }
36         bool validFinal[] = {0, 0, 0, 1, 0, 0, 1, 1, 1};
37         return validFinal[last];
38     }
39 };

我用JAVA改写的：

 1 public class Solution {
 2     public boolean isNumber(String s) {
 3         int[][] transitionTable={
 4             /* 　0   1   2   3   4   5  */
 5               　{0,  1,  2,  3,  4,  0}, // 0: INVALID
 6              　 {0,  1,  2,  3,  4,  0}, // 1: SPACE
 7                 {0,  0,  0,  3,  4,  0}, // 2: SIGN
 8                 {0,  6,  0,  3,  7,  5}, // 3: DIGIT
 9                 {0,  0,  0,  7,  0,  0}, // 4: DOT
10                 {0,  0,  2,  8,  0,  0}, // 5: EXPONENT
11                 {0,  6,  0,  0,  0,  0}, // 6: END WITH SPACE
12                 {0,  6,  0,  7,  0,  5}, // 7: DOT AND DIGIT
13                 {0,  6,  0,  8,  0,  0}, // 8: END WITH SPACE OR DIGIT           
14         };
15         int last=0;　　　　　　//invalid
16         for(int i=0;i<s.length();i++){
17             int state=0;　　　//invalid
18             if(s.charAt(i)==' '){
19                 state=1;　　　//space
20             }else if(Character.isDigit(s.charAt(i))){
21                 state=3;　　　//digit
22             }else if(s.charAt(i)=='+'||s.charAt(i)=='-'){
23                 state=2;　　　//SIGN
24             }else if(s.charAt(i)=='.'){
25                 state=4;　　　//DOT
26             }else if(s.charAt(i)=='e'||s.charAt(i)=='E'){
27                 state=5;　　　//EXPONENT
28             }
29             last=transitionTable[last][state];
30             if(last==0) return false;
31         }
32         if(last==3||last==6||last==7||last==8)　　//最后结束状态为3,6,7,8时表示该string确实是一个number
33             return true;
34         else
35             return false;
36         
37     }
38 }

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其实很想像AnnieKim一样用enum的。。。但是自己enum太不熟了。

　　public static void main(String[] args) {
        // TODO Auto-generated method stub
        Main so = new Main();
        InputType a = InputType.SIGN;
        
        System.out.println(a.ordinal());　　//2
    }

    public enum InputType {
        INVALID, SPACE, SIGN, DIGIT, DOT, EXPONENT
    }

通过ordinal函数得到该数的序列号。

但是想要将int值如2转换成InputType的SIGN，要自己写个函数来实现。