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2014
11-19

LeetCode-Valid Number[字符串]

Valid Number

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.

标签: Math String
分析

细节实现题。

本题的功能与标准库中的{strtod()}功能类似。

代码1

// LeetCode, Valid Number
// @author 龚陆安 (http://weibo.com/luangong)
// finite automata,时间复杂度O(n),空间复杂度O(n)
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
        };
        const int transitionTable[][NUM_INPUTS] = {
                -1, 0, 3, 1, 2, -1, // next states for state 0
                -1, 8, -1, 1, 4, 5,     // next states for state 1
                -1, -1, -1, 4, -1, -1,     // next states for state 2
                -1, -1, -1, 1, 2, -1,     // next states for state 3
                -1, 8, -1, 4, -1, 5,     // next states for state 4
                -1, -1, 6, 7, -1, -1,     // next states for state 5
                -1, -1, -1, 7, -1, -1,     // next states for state 6
                -1, 8, -1, 7, -1, -1,     // next states for state 7
                -1, 8, -1, -1, -1, -1,     // next states for state 8
                };

        int state = 0;
        for (; *s != '\0'; ++s) {
            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;
        }
        // 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;

    }
};

代码2

// LeetCode, Valid Number
// @author 连城 (http://weibo.com/lianchengzju)
// 偷懒,直接用 strtod(),时间复杂度O(n)
class Solution {
public:
    bool isNumber (char const* s) {
        char* endptr;
        strtod (s, &endptr);

        if (endptr == s) return false;

        for (; *endptr; ++endptr)
            if (!isspace (*endptr)) return false;

        return true;
    }
};