leetcode valid Number的2种状态定义

把一个valid number看做一个顺序序列，把各个关键位置应为状态，比如开始空白部分，符号后，整数部分，小数部分，指数部分等

状态转移是递增的，往下走不会回退。这种定义类似KMP和正则的NFA的状态定义，状态就是处在pattern中的位置。

下的程序有bug, "."会被当作valid，懒得改了，需要调整下有些状态定义，但是不出“状态就是pattern中的位置”

bool isNumber(string s) {enum { EMPTY, AFTER_SIGN, IN_INTEGER, AFTER_INTEGER, AFTER_POINT, AFTER_POINT2, IN_FRACTION, AFTER_FRACTION, AFTER_E, AFTER_EXP_SIGN, IN_EXP_INTEGER, AFTER_EXP_INTEGER } state;state = EMPTY;for (char &c : s) {if (c == ' ') {if (state == IN_INTEGER || state == AFTER_INTEGER ) state = AFTER_INTEGER;else if (state == AFTER_POINT) state = AFTER_POINT2;else if (state == IN_EXP_INTEGER || state == AFTER_EXP_INTEGER) state = AFTER_EXP_INTEGER;else if (state == IN_FRACTION) state = AFTER_FRACTION;else if (state != EMPTY && state != AFTER_POINT2 && state != AFTER_FRACTION) return false;}else if (c == '+' || c == '-') {if (state != EMPTY && state != AFTER_E) return false;if (state == AFTER_E) state = AFTER_EXP_SIGN;else state = AFTER_SIGN;}else if (isdigit(c)) {if (state == AFTER_INTEGER || state == AFTER_POINT2) return false;if (state == AFTER_POINT || state == IN_FRACTION) state = IN_FRACTION;else if (state == AFTER_E) state = IN_EXP_INTEGER;else state = IN_INTEGER;}else if (c == '.') {if (state != IN_INTEGER && state != EMPTY && state != AFTER_SIGN) return false;state = AFTER_POINT;}else if (c == 'e' || c == 'E') {if (state != IN_FRACTION && state != IN_INTEGER && state != AFTER_POINT) return false;state = AFTER_E;}else return false;}return state == IN_INTEGER || state == AFTER_INTEGER || state == AFTER_POINT || state == AFTER_POINT2|| state == IN_FRACTION || state == AFTER_FRACTION|| state == IN_EXP_INTEGER || state == AFTER_EXP_INTEGER;}

还有一种状态定义：输入序列到当前为止，是否是valid number

def isNumber(self, s):i, n = 0, len(s)while i < n and s[i]==' ': i += 1if i < n and s[i] in '+-': i += 1isNum = Falsewhile i < n and s[i].isdigit():i += 1isNum = Trueif i < n and s[i] =='.':i += 1while i < n and s[i].isdigit():i += 1isNum = Trueif i < n and isNum and s[i] == 'e':i += 1isNum = Falseif i < n and s[i] in '+-': i += 1while i < n and s[i].isdigit():i += 1isNum = Truewhile i < n and s[i] == ' ': i += 1return isNum and i == n