[LeetCode] 65. Valid Number 验证数字

2021年09月15日 阅读数:1
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Validate if a given string can be interpreted as a decimal number.html

Some examples:
"0" => true
" 0.1 " => true
"abc" => false
"1 a" => false
"2e10" => true
" -90e3   " => true
" 1e" => false
"e3" => false
" 6e-1" => true
" 99e2.5 " => false
"53.5e93" => true
" --6 " => false
"-+3" => false
"95a54e53" => falsejava

Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. However, here is a list of characters that can be in a valid decimal number:python

  • Numbers 0-9
  • Exponent - "e"
  • Positive/negative sign - "+"/"-"
  • Decimal point - "."

Of course, the context of these characters also matters in the input.git

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.面试

这道题须要考虑的状况很是多,OJ的成功率很低,估计面试不会出此题的。post

Java:ui

public boolean isNumber(String s) {
    s = s.trim();
    
    boolean pointSeen = false;
    boolean eSeen = false;
    boolean numberSeen = false;
    boolean numberAfterE = true;
    for(int i=0; i<s.length(); i++) {
        if('0' <= s.charAt(i) && s.charAt(i) <= '9') {
            numberSeen = true;
            numberAfterE = true;
        } else if(s.charAt(i) == '.') {
            if(eSeen || pointSeen) {
                return false;
            }
            pointSeen = true;
        } else if(s.charAt(i) == 'e') {
            if(eSeen || !numberSeen) {
                return false;
            }
            numberAfterE = false;
            eSeen = true;
        } else if(s.charAt(i) == '-' || s.charAt(i) == '+') {
            if(i != 0 && s.charAt(i-1) != 'e') {
                return false;
            }
        } else {
            return false;
        }
    }
    
    return numberSeen && numberAfterE;
}

Python:url

class Solution:
    # @param s, a string
    # @return a boolean
    # @finite automation
    def isNumber(self, s):
        INVALID=0; SPACE=1; SIGN=2; DIGIT=3; DOT=4; EXPONENT=5;
        #0invalid,1space,2sign,3digit,4dot,5exponent,6num_inputs
        transitionTable=[[-1,  0,  3,  1,  2, -1],    #0 no input or just spaces 
                         [-1,  8, -1,  1,  4,  5],    #1 input is digits 
                         [-1, -1, -1,  4, -1, -1],    #2 no digits in front just Dot 
                         [-1, -1, -1,  1,  2, -1],    #3 sign 
                         [-1,  8, -1,  4, -1,  5],    #4 digits and dot in front 
                         [-1, -1,  6,  7, -1, -1],    #5 input 'e' or 'E' 
                         [-1, -1, -1,  7, -1, -1],    #6 after 'e' input sign 
                         [-1,  8, -1,  7, -1, -1],    #7 after 'e' input digits 
                         [-1,  8, -1, -1, -1, -1]]    #8 after valid input input space
        state=0; i=0
        while i<len(s):
            inputtype = INVALID
            if s[i]==' ': inputtype=SPACE
            elif s[i]=='-' or s[i]=='+': inputtype=SIGN
            elif s[i] in '0123456789': inputtype=DIGIT
            elif s[i]=='.': inputtype=DOT
            elif s[i]=='e' or s[i]=='E': inputtype=EXPONENT
            
            state=transitionTable[state][inputtype]
            if state==-1: return False
            else: i+=1
        return state == 1 or state == 4 or state == 7 or state == 8
                 

C++:spa

class Solution {
public:
    bool isNumber(string s) {
        bool num = false, numAfterE = true, dot = false, exp = false, sign = false;
        int n = s.size();
        for (int i = 0; i < n; ++i) {
            if (s[i] == ' ') {
                if (i < n - 1 && s[i + 1] != ' ' && (num || dot || exp || sign)) return false;
            } else if (s[i] == '+' || s[i] == '-') {
                if (i > 0 && s[i - 1] != 'e' && s[i - 1] != ' ') return false;
                sign = true;
            } else if (s[i] >= '0' && s[i] <= '9') {
                num = true;
                numAfterE = true;
            } else if (s[i] == '.') {
                if (dot || exp) return false;
                dot = true;
            } else if (s[i] == 'e') {
                if (exp || !num) return false;
                exp = true;
                numAfterE = false;
            } else return false;
        }
        return num && numAfterE;
    }
};

  

 

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