LeetCode OJ:Valid Number

网上有人的实现方法很不错，加了些注释

class Solution {
public:
    bool isNumber(const char *s) {
    enum InputType {
        INVALID,    // 0
        SPACE,      // 1
        SIGN,       // 2
        DIGIT,      // 3
        DOT,        // 4
        EXPONENT,   // 5
        NUM_INPUTS  // 6
    };
    int transitionTable[][NUM_INPUTS] = {
        -1,  0,  3,  1,  2, -1,     // next states for state 0：第一个字符
        -1,  8, -1,  1,  4,  5,     // next states for state 1：小数点前的数字
        -1, -1, -1,  4, -1, -1,     // next states for state 2：小数点
        -1, -1, -1,  1,  2, -1,     // next states for state 3：正负号
        -1,  8, -1,  4, -1,  5,     // next states for state 4：小数点后的数字或者是数字后的小数点
        -1, -1,  6,  7, -1, -1,     // next states for state 5：e
        -1, -1, -1,  7, -1, -1,     // next states for state 6：e后面的正负号
        -1,  8, -1,  7, -1, -1,     // next states for state 7：e后面的数字
        -1,  8, -1, -1, -1, -1,     // next states for state 8：空格
    };

    int state = 0;
    while (*s != ‘\0‘) {
        InputType inputType = INVALID;
        if (isspace(*s))
            inputType = SPACE;
        else if (*s == ‘+‘ || *s == ‘-‘)
            inputType = SIGN;
        else if (isdigit(*s))
            inputType = DIGIT;
        else if (*s == ‘.‘)
            inputType = DOT;
        else if (*s == ‘e‘ || *s == ‘E‘)
            inputType = EXPONENT;
    
        // Get next state from current state and input symbol
        state = transitionTable[state][inputType];
    
        // Invalid input
        if (state == -1) return false;
        else ++s;
    }
    // If the current state belongs to one of the accepting (final) states,
    // then the number is valid
    return state == 1 || state == 4 || state == 7 || state == 8;
    
    }
};

LeetCode OJ:Valid Number

原文：http://blog.csdn.net/starcuan/article/details/18845893