In this post, we are going to solve the Valid Number Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Problem
A valid number can be split up into these components (in order):
- A decimal number or an integer.
- (Optional) An
'e'
or'E'
, followed by an integer.
A decimal number can be split up into these components (in order):
- (Optional) A sign character (either
'+'
or'-'
). - One of the following formats:
An integer can be split up into these components (in order):
- (Optional) A sign character (either
'+'
or'-'
). - 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'.'
.
Now, let’s see the leetcode solution of Valid Number Leetcode Solution.
Valid Number Leetcode Solution in Python
class Solution: def isNumber(self, s: str) -> bool: s = s.strip() if not s: return False seenNum = False seenDot = False seenE = False for i, c in enumerate(s): if c == '.': if seenDot or seenE: return False seenDot = True elif c == 'e' or c == 'E': if seenE or not seenNum: return False seenE = True seenNum = False elif c in '+-': if i > 0 and s[i - 1] != 'e': return False seenNum = False else: if not c.isdigit(): return False seenNum = True return seenNum
Valid Number Leetcode Solution in CPP
class Solution { public: bool isNumber(string s) { trim(s); if (s.empty()) return false; bool seenNum = false; bool seenDot = false; bool seenE = false; for (int i = 0; i < s.length(); ++i) { switch (s[i]) { case '.': if (seenDot || seenE) return false; seenDot = true; break; case 'e': case 'E': if (seenE || !seenNum) return false; seenE = true; seenNum = false; break; case '+': case '-': if (i > 0 && s[i - 1] != 'e') return false; seenNum = false; break; default: if (!isdigit(s[i])) return false; seenNum = true; } } return seenNum; } private: void trim(string& s) { s.erase(0, s.find_first_not_of(' ')); s.erase(s.find_last_not_of(' ') + 1); } };
Valid Number Leetcode Solution in Java
class Solution { public boolean isNumber(String s) { s = s.trim(); if (s.isEmpty()) return false; boolean seenNum = false; boolean seenDot = false; boolean seenE = false; for (int i = 0; i < s.length(); ++i) { switch (s.charAt(i)) { case '.': if (seenDot || seenE) return false; seenDot = true; break; case 'e': case 'E': if (seenE || !seenNum) return false; seenE = true; seenNum = false; break; case '+': case '-': if (i > 0 && s.charAt(i - 1) != 'e') return false; seenNum = false; break; default: if (!Character.isDigit(s.charAt(i))) return false; seenNum = true; } } return seenNum; } }
Note: This problem Valid Number is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.
NEXT: Plus One Leetcode