LeetCode - Valid Number | Darren's Blog

Validate if a given string is numeric.
Some examples:
"0" => true
" 0.1 " => true
"abc" => false
"1 a" => false
"2e10" => true

public boolean isNumber(String s) {

        s = s.trim();       // Get rid of leading and trailing whitespaces

        int n = s.length();

        if (n == 0)

            return false;

        boolean isFractional = false;   // Indicate appearance of '.'

        boolean isExponential = false;  // Indicate appearance of 'e/E'

        boolean valid = false;          // Indicate whether preceding digits are valid

        boolean expoBefore = false;     // Indicate whether the preceding digit is 'e/E'

        boolean validFrac = true;       // Indicate whether the number is a vaild fraction (true by default)

        boolean validExpo = true;       // Indicate whether the number is a valid exponential (true by default)

        int i = 0;

        if (s.charAt(0) == '+' || s.charAt(0) == '-')   // Consider sign

            i = 1;

        // Process each digit in sequence

        for (; i < n; i++) {

            char c = s.charAt(i);

            if (c == '+' || c == '-') {     // +/- is valid only after e/E

                if (!expoBefore)

                    return false;

                expoBefore = false;

                valid = false;

            } else if (c == 'e' || c == 'E') {  // Only one e/E is valid; preceding digits should be valid

                if (isExponential || !valid)

                    return false;

                isExponential = true;

                valid = false;

                expoBefore = true;

                validExpo = false;

            } else if (c == '.') {  // Only one '.' is valid; cannot appear as an exponential

                if (isFractional || isExponential)

                    return false;

                isFractional = true;

                if (!valid)     // Must have fractional part

                    validFrac = false;

            } else if (c >= '0' && c <='9') {   // Regular digits

                valid = true;

                expoBefore = false;

                validFrac = true;

                validExpo = true;

            } else {

                return false;

            }

        }

        // After reaching the end, make sure the number is indeed valid

        if (!valid || !validFrac || !validExpo)

            return false;

        else

            return true;

    }

Use DFA(Deterministic Finite Automata)
Code from http://jelices.blogspot.com/2013/12/leetcode-valid-number.html
We accept the following regular grammar, shown as UNIX regular expression by simplicity:
\s*[\+\-]?(([0-9]+(\.[0-9]*)?)|\.[0-9]+)([eE][+-]?[0-9]+)?\s*
where \s represents [ \t].

This regular expression is accepted by the underneath DFA(Deterministic Finite Automata):

public boolean isNumber(String s) 
    {
        int state = 0;
        for(int i=0; i<s.length();i++)
        {
            state=getNextState(state, s.charAt(i));
            if (state==-1)
                return false;
        }
        state = getNextState(state, ' ');
        return state == 8;
    }
    
    private int getNextState(int state, char c)
    {
        int[][] transitionTable = { {0,2,1,3,-1,-1},
                                    {-1,2,-1,3,-1,-1},
                                    {8,2,-1,4,5,-1},
                                    {-1,4,-1,-1,-1,-1},
                                    {8,4,-1,-1,5,-1},
                                    {-1,7,6,-1,-1,-1},
                                    {-1,7,-1,-1,-1,-1},
                                    {8,7,-1,-1,-1,-1},
                                    {8,-1,-1,-1,-1,-1}};
        return transitionTable[state][getSymbol(c)];
    }
    
    private int getSymbol(char c)
    {
        switch(c)
        {
            case ' ': case '\t': return 0;
            case '0': case '1': case '2':case '3': case '4': case '5': case '6': case '7': case '8': case '9': return 1;
            case '+': case '-': return 2;
            case '.': return 3;
            case 'E': case 'e': return 4;
            default: return 5;
        }
    }

Use Regular Expression:
Code from http://www.cnblogs.com/midnightcat/p/3364431.html

Pattern p = Pattern.compile("^[\\+\\-]?((\\d+(\\.\\d*)?)|(\\.\\d+))(e[\\+\\-]?\\d+)?$");
    public boolean isNumber(String s) {
        String ss = s.trim();
        return p.matcher(ss).matches();
    }

Also refer to http://blog.csdn.net/ithomer/article/details/8832300
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