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 <= 20sconsists 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
