Last updated on October 9th, 2024 at 10:34 pm

Here, We see Valid Number LeetCode Solution. This Leetcode problem done in many programming language like C++, Java, JavaScript, Python etc. with different approaches.

List of all LeetCode Solution

Valid Number LeetCode Solution

Problem Statement

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

1. Valid Number Leetcode Solution C++

class Solution {
public:
    bool isNumber(string s) {
        bool digitSeen=false, dotSeen=false, eSeen=false;
        int plusMinus=0,n=s.length();
        for(int i=0; i<n; i++)
        {
            if(s[i]-'0'>=0 && s[i]-'0'<=9)
                digitSeen=true;
            else if(s[i]=='+' || s[i]=='-')
            {
                if(plusMinus==2 || (i>0 && (s[i-1]!='e' && s[i-1]!='E')) || i==n-1) return false;
                plusMinus++;
            }
            else if(s[i]=='e' || s[i]=='E')
            {
                if(eSeen || !digitSeen || i==n-1) return false;
                eSeen=true;
            }
            else if(s[i]=='.')
            {
                if(eSeen || dotSeen || (i==n-1 && !digitSeen)) return false;
                dotSeen=true;
            }
            else
              return false;
        }
        return true;        
    }
};

2. Valid Number Leetcode Solution Java

class Solution {
    public boolean isNumber(String s) {
        boolean num = false, exp = false, sign = false, dec = false;
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c >= '0' && c <= '9') num = true ;    
            else if (c == 'e' || c == 'E')
                if (exp || !num) return false;
                else {
                    exp = true;
                    sign = false;
                    num = false;
                    dec = false;
                }
            else if (c == '+' || c == '-')
                if (sign || num || dec) return false;
                else sign = true;
            else if (c == '.')
                if (dec || exp) return false;
                else dec = true;
            else return false;
        }
        return num;        
    }
}

3. Valid Number Leetcode Solution JavaScript

var isNumber = function(s) {
    let exp = false, sign = false, num = false, dec = false
    for (let c of s)
        if (c >= '0' && c <= '9') num = true     
        else if (c === 'e' || c === 'E')
            if (exp || !num) return false
            else exp = true, sign = false, num = false, dec = false
        else if (c === '+' || c === '-')
            if (sign || num || dec) return false
            else sign = true
        else if (c === '.')
            if (dec || exp) return false
            else dec = true
        else return false
    return num    
};

4. Valid Number Leetcode Solution Python

class Solution(object):
    def isNumber(self, s):
        try:
            if 'inf' in s.lower() or s.isalpha():
                return False
            if float(s) or float(s) >= 0:
                return True
        except:
            return False