[LeetCode]Valid Number

最新推荐文章于 2020-03-20 11:15:29 发布

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最新推荐文章于 2020-03-20 11:15:29 发布

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分类专栏： LeetCode

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本文介绍了一种使用有限状态机（FSM）来判断输入字符串是否能被解析为有效数字的方法。通过两个不同的实现展示了如何构建状态转换表，并详细解释了每个状态的意义及其在遇到不同字符时的状态转移逻辑。

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class Solution {
//finite state machine
//status 1 => 小数点前面的部分
//status 2 => 小数点后到'e'前面的部分
//status 3 => e后面的科学计数法部分
//should practice couple times more
public:
	bool isNumber(const char *s) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		bool bIntNum = false;
		int nCurState = 1;
		while (true)
		{
			if (nCurState == 1)
			{
				while (*s == ' ')
					s++;
				if (*s == '+' || *s == '-')
					s++;
				if (*s >= '0' && *s <= '9')
				{
					while (*s >= '0' && *s <= '9')
						s++;
					bIntNum = true;
				}

				if (*s == '.' /*&& bIntNum*/) //.34 true
				{
					nCurState = 2;
					s++;
					continue;
				}
				else if (*s == 'e' && bIntNum)//go to state 3, there must have bIntNum before
				{
					nCurState = 3;
					s++;
					continue;
				}
				while (*s == ' ')
					s++;
				return (bIntNum && *s == 0);
			}
			else if (nCurState == 2)//1. true
			{
				//bIntNum = false;
				if (*s >= '0' && *s <= '9')
				{
					while (*s >= '0' && *s <= '9')
						s++;
					bIntNum = true;
				}
				if (*s == 'e' && bIntNum)//go to state 3, there must have bIntNum before
				{
					nCurState = 3;
					s++;
					continue;
				}
				while (*s == ' ')
					s++;
				return (bIntNum && *s == 0);//if there is no e, then there must have bIntNum, such as ".23" or "1."
			}
			else//nCurState 3 need judge if there is bIntNum again, because "1e" is illegal
			{
				bIntNum = false;
				if (*s == '+' || *s == '-')
					s++;
				if (*s >= '0' && *s <= '9')
				{
					while (*s >= '0' && *s <= '9')
						s++;
					bIntNum = true;
				}
				while (*s == ' ')
					s++;
				return (*s == 0 && bIntNum);
			}
		}
	}
};

second time

class Solution {
//finite automata
public:
    vector<vector<int> > transTable;
    int inputType(char a)
    {
        if(a == ' ') return 0;
        else if(a == '+' || a == '-') return 1;
        else if(a >= '0' && a <= '9') return 2;
        else if(a == '.') return 3;
        else if(a == 'e') return 4;
        else return 5;//illegal input
    }
    void initTransTable()
    {
        transTable.resize(9, vector<int>(5, -1));
        transTable[0][0] = 0;
        transTable[0][1] = 1;
        transTable[0][2] = 2;
        transTable[0][3] = 7;
        transTable[7][2] = 3;
        transTable[1][2] = 2;
        transTable[1][3] = 7;
        transTable[2][2] = 2;
        transTable[2][0] = 6;
        transTable[2][3] = 3;
        transTable[2][4] = 4;
        transTable[3][2] = 3;
        transTable[3][4] = 4;
        transTable[3][0] = 6;
        transTable[4][1] = 8;
        transTable[8][2] = 5;
        transTable[4][2] = 5;
        transTable[5][2] = 5;
        transTable[5][0] = 6;
        transTable[6][0] = 6;
    }
    bool isNumber(const char *s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(s == NULL) return false;
        initTransTable();
        int curState = 0;
        while(*s != '\0')
        {
            int curInput = inputType(*s);
            if(curInput == 5) return false;
            curState = transTable[curState][curInput];
            if(curState == -1) return false;
            s++;
        }
        if(curState == 2 || curState == 3 || curState == 5 || curState == 6) return true;
        else return false; 
    }
};