[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。

public class Solution {
    public boolean isNumber(String s) {
        s=s.trim();
        s=s.replaceAll("e|E", "/");
        s=s.replaceAll("[a-zA-Z]", "x");
        s=s.replaceAll("/", "e");
    
        try {
            Double.parseDouble(s);
        } catch (Exception e) {
            // TODO: handle exception
        
            return false;
        }
    
        return true;
    }
}
 

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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

class Solution {
public:
    bool isNumber(const char *s) {
        enum InputType {INVALID, SPACE, SIGN, DIGIT, DOT, EXPONENT};
        int transitionTable[][SPACEEND] = 
        { /* 0   1   2   3   4   5  */
             0,  1,  2,  3,  4,  0, // 0: INVALID
             0,  1,  2,  3,  4,  0, // 1: SPACE
             0,  0,  0,  3,  4,  0, // 2: SIGN
             0,  6,  0,  3,  7,  5, // 3: DIGIT
             0,  0,  0,  7,  0,  0, // 4: DOT
             0,  0,  2,  8,  0,  0, // 5: EXPONENT
             0,  6,  0,  0,  0,  0, // 6: END WITH SPACE
             0,  6,  0,  7,  0,  5, // 7: DOT AND DIGIT
             0,  6,  0,  8,  0,  0, // 8: END WITH SPACE OR DIGIT
        };
        
        InputType last = INVALID;
        while (*s != '')
        {
            InputType state = INVALID;
            if (*s == ' ')
                state = SPACE;
            else if (isdigit(*s))
                state = DIGIT;
            else if (*s == '+' || *s == '-')
                state = SIGN;
            else if (*s == 'e')
                state = EXPONENT;
            else if (*s == '.')
                state = DOT;
            last = (InputType) transitionTable[last][state];
            if (last == INVALID) return false;
            s++;
        }
        bool validFinal[] = {0, 0, 0, 1, 0, 0, 1, 1, 1};
        return validFinal[last];
    }
};

 我用JAVA改写的：

public class Solution {
    public boolean isNumber(String s) {
        int[][] transitionTable={
            /* 　0   1   2   3   4   5  */
              　{0,  1,  2,  3,  4,  0}, // 0: INVALID
             　 {0,  1,  2,  3,  4,  0}, // 1: SPACE
                {0,  0,  0,  3,  4,  0}, // 2: SIGN
                {0,  6,  0,  3,  7,  5}, // 3: DIGIT
                {0,  0,  0,  7,  0,  0}, // 4: DOT
                {0,  0,  2,  8,  0,  0}, // 5: EXPONENT
                {0,  6,  0,  0,  0,  0}, // 6: END WITH SPACE
                {0,  6,  0,  7,  0,  5}, // 7: DOT AND DIGIT
                {0,  6,  0,  8,  0,  0}, // 8: END WITH SPACE OR DIGIT           
        };
        int last=0;　　　　　　//invalid
        for(int i=0;i<s.length();i++){
            int state=0;　　　//invalid
            if(s.charAt(i)==' '){
                state=1;　　　//space
            }else if(Character.isDigit(s.charAt(i))){
                state=3;　　　//digit
            }else if(s.charAt(i)=='+'||s.charAt(i)=='-'){
                state=2;　　　//SIGN
            }else if(s.charAt(i)=='.'){
                state=4;　　　//DOT
            }else if(s.charAt(i)=='e'||s.charAt(i)=='E'){
                state=5;　　　//EXPONENT
            }
            last=transitionTable[last][state];
            if(last==0) return false;
        }
        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
            return true;
        else
            return false;
        
    }
}

--------------------------------------------------------------

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。