Valid Number LeetCode Solution

Problem – Valid Number LeetCode Solution

A valid number can be split up into these components (in order):

1. A decimal number or an integer.
2. (Optional) An 'e' or 'E', followed by an integer.

A decimal number can be split up into these components (in order):

1. (Optional) A sign character (either '+' or '-').
2. One of the following formats:
   1. One or more digits, followed by a dot '.'.
   2. One or more digits, followed by a dot '.', followed by one or more digits.
   3. A dot '.', followed by one or more digits.

An integer can be split up into these components (in order):

1. (Optional) A sign character (either '+' or '-').
2. One or more digits.

For example, all the following are valid numbers: ["2", "0089", "-0.1", "+3.14", "4.", "-.9", "2e10", "-90E3", "3e+7", "+6e-1", "53.5e93", "-123.456e789"], while the following are not valid numbers: ["abc", "1a", "1e", "e3", "99e2.5", "--6", "-+3", "95a54e53"].

Given a string s, return true if s is a valid number.

Example 1:

Input: s = "0"
Output: true

Example 2:

Input: s = "e"
Output: false

Example 3:

Input: s = "."
Output: false

Constraints:

• 1 <= s.length <= 20
• s consists of only English letters (both uppercase and lowercase), digits (0-9), plus '+', minus '-', or dot '.'.

Valid Number LeetCode Solution in Java

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;
}

Valid Number LeetCode Solution in Python

class Solution(object):
  def isNumber(self, s):
      """
      :type s: str
      :rtype: bool
      """
      #define a DFA
      state = [{}, 
              {'blank': 1, 'sign': 2, 'digit':3, '.':4}, 
              {'digit':3, '.':4},
              {'digit':3, '.':5, 'e':6, 'blank':9},
              {'digit':5},
              {'digit':5, 'e':6, 'blank':9},
              {'sign':7, 'digit':8},
              {'digit':8},
              {'digit':8, 'blank':9},
              {'blank':9}]
      currentState = 1
      for c in s:
          if c >= '0' and c <= '9':
              c = 'digit'
          if c == ' ':
              c = 'blank'
          if c in ['+', '-']:
              c = 'sign'
          if c not in state[currentState].keys():
              return False
          currentState = state[currentState][c]
      if currentState not in [3,5,8,9]:
          return False
      return True

Valid Number LeetCode Solution in C++

bool isNumber(const char *s) 
{
    int i = 0;
    
    // skip the whilespaces
    for(; s[i] == ' '; i++) {}
    
    // check the significand
    if(s[i] == '+' || s[i] == '-') i++; // skip the sign if exist
    
    int n_nm, n_pt;
    for(n_nm=0, n_pt=0; (s[i]<='9' && s[i]>='0') || s[i]=='.'; i++)
        s[i] == '.' ? n_pt++:n_nm++;       
    if(n_pt>1 || n_nm<1) // no more than one point, at least one digit
        return false;
    
    // check the exponent if exist
    if(s[i] == 'e') {
        i++;
        if(s[i] == '+' || s[i] == '-') i++; // skip the sign
        
        int n_nm = 0;
        for(; s[i]>='0' && s[i]<='9'; i++, n_nm++) {}
        if(n_nm<1)
            return false;
    }
    
    // skip the trailing whitespaces
    for(; s[i] == ' '; i++) {}
    
    return s[i]==0;  // must reach the ending 0 of the string
}

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