leetcode-65-面试题20-表示数值的字符串

Posted 2021-03-01 oldby

题目描述：

 

 方法一：逻辑判断法：O(N)

class Solution:
    def isNumber(self, s: str) -> bool:
        s = s.strip()
        met_dot = met_e = met_digit = False
        for i, char in enumerate(s):
            if char in (‘+‘, ‘-‘):
                if i > 0 and s[i-1] != ‘e‘ and s[i-1] != ‘E‘:
                    return False
            elif char == ‘.‘:
                if met_dot or met_e: 
                    return False
                met_dot = True
            elif char == ‘e‘ or char == ‘E‘:
                if met_e or not met_digit:
                    return False
                met_e, met_digit = True, False # e后必须接，所以这时重置met_digit为False,以免e为最后一个char
            elif char.isdigit():
                met_digit = True
            else:
                return False
        return met_digit

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方法二;正则表达式

class Solution:
    def isNumber(self, s: str) -> bool:
        p = re.compile(r‘^[+-]?(.d+|d+.?d*)([eE][+-]?d+)?$‘)
        return bool(p.match(s.strip()))

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方法三：try/except法

class Solution:
    def isNumber(self, s: str) -> bool:
        try:
            float(s)
            return True
        except:
            return False

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方法四：确定性有限自动机

class Solution(object):
    def isNumber(self, s):
        """
        :type s: str
        :rtype: bool
        """
        #define DFA state transition tables
        states = [{},
                 # State (1) - initial state (scan ahead thru blanks)
                 {‘blank‘: 1, ‘sign‘: 2, ‘digit‘:3, ‘.‘:4},
                 # State (2) - found sign (expect digit/dot)
                 {‘digit‘:3, ‘.‘:4},
                 # State (3) - digit consumer (loop until non-digit)
                 {‘digit‘:3, ‘.‘:5, ‘e‘:6, ‘blank‘:9},
                 # State (4) - found dot (only a digit is valid)
                 {‘digit‘:5},
                 # State (5) - after dot (expect digits, e, or end of valid input)
                 {‘digit‘:5, ‘e‘:6, ‘blank‘:9},
                 # State (6) - found ‘e‘ (only a sign or digit valid)
                 {‘sign‘:7, ‘digit‘:8},
                 # State (7) - sign after ‘e‘ (only digit)
                 {‘digit‘:8},
                 # State (8) - digit after ‘e‘ (expect digits or end of valid input)
                 {‘digit‘:8, ‘blank‘:9},
                 # State (9) - Terminal state (fail if non-blank found)
                 {‘blank‘:9}]
        currentState = 1
        for c in s:
            # If char c is of a known class set it to the class name
            if c in ‘0123456789‘:
                c = ‘digit‘
            elif c in ‘ 	
‘:
                c = ‘blank‘
            elif c in ‘+-‘:
                c = ‘sign‘
            # If char/class is not in our state transition table it is invalid input
            if c not in states[currentState]:
                return False
            # State transition
            currentState = states[currentState][c]
        # The only valid terminal states are end on digit, after dot, digit after e, or white space after valid input
        if currentState not in [3,5,8,9]:
            return False
        return True