[LeetCode] 65. Valid Number 验证数字

Validate if a given string can be interpreted as a decimal number.

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" => false

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:

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

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

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.

这道题需要考虑的情况非常多，OJ的成功率很低，估计面试不会出此题的。

Java:

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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:

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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++:

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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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