# 65. Valid Number

## Description

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 '.'.

## Solutions

Solution 1: Case Discussion

First, we check if the string starts with a positive or negative sign. If it does, we move the pointer $i$ one step forward. If the pointer $i$ has reached the end of the string at this point, it means the string only contains a positive or negative sign, so we return false.

If the character pointed to by the current pointer $i$ is a decimal point, and there is no number after the decimal point, or if there is an e or E after the decimal point, we return false.

Next, we use two variables $dot$ and $e$ to record the number of decimal points and e or E respectively.

We use pointer $j$ to point to the current character:

• If the current character is a decimal point, and a decimal point or e or E has appeared before, return false. Otherwise, we increment $dot$ by one;
• If the current character is e or E, and e or E has appeared before, or if the current character is at the beginning or end of the string, return false. Otherwise, we increment $e$ by one; then check if the next character is a positive or negative sign, if it is, move the pointer $j$ one step forward. If the pointer $j$ has reached the end of the string at this point, return false;
• If the current character is not a number, return false.

After traversing the string, return true.

The time complexity is $O(n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the string.

• class Solution {
public boolean isNumber(String s) {
int n = s.length();
int i = 0;
if (s.charAt(i) == '+' || s.charAt(i) == '-') {
++i;
}
if (i == n) {
return false;
}
if (s.charAt(i) == '.'
&& (i + 1 == n || s.charAt(i + 1) == 'e' || s.charAt(i + 1) == 'E')) {
return false;
}
int dot = 0, e = 0;
for (int j = i; j < n; ++j) {
if (s.charAt(j) == '.') {
if (e > 0 || dot > 0) {
return false;
}
++dot;
} else if (s.charAt(j) == 'e' || s.charAt(j) == 'E') {
if (e > 0 || j == i || j == n - 1) {
return false;
}
++e;
if (s.charAt(j + 1) == '+' || s.charAt(j + 1) == '-') {
if (++j == n - 1) {
return false;
}
}
} else if (s.charAt(j) < '0' || s.charAt(j) > '9') {
return false;
}
}
return true;
}
}

• class Solution {
public:
bool isNumber(string s) {
int n = s.size();
int i = 0;
if (s[i] == '+' || s[i] == '-') ++i;
if (i == n) return false;
if (s[i] == '.' && (i + 1 == n || s[i + 1] == 'e' || s[i + 1] == 'E')) return false;
int dot = 0, e = 0;
for (int j = i; j < n; ++j) {
if (s[j] == '.') {
if (e || dot) return false;
++dot;
} else if (s[j] == 'e' || s[j] == 'E') {
if (e || j == i || j == n - 1) return false;
++e;
if (s[j + 1] == '+' || s[j + 1] == '-') {
if (++j == n - 1) return false;
}
} else if (s[j] < '0' || s[j] > '9')
return false;
}
return true;
}
};

• class Solution:
def isNumber(self, s: str) -> bool:
n = len(s)
i = 0
if s[i] in '+-':
i += 1
if i == n:
return False
if s[i] == '.' and (i + 1 == n or s[i + 1] in 'eE'):
return False
dot = e = 0
j = i
while j < n:
if s[j] == '.':
if e or dot:
return False
dot += 1
elif s[j] in 'eE':
if e or j == i or j == n - 1:
return False
e += 1
if s[j + 1] in '+-':
j += 1
if j == n - 1:
return False
elif not s[j].isnumeric():
return False
j += 1
return True


• func isNumber(s string) bool {
i, n := 0, len(s)
if s[i] == '+' || s[i] == '-' {
i++
}
if i == n {
return false
}
if s[i] == '.' && (i+1 == n || s[i+1] == 'e' || s[i+1] == 'E') {
return false
}
var dot, e int
for j := i; j < n; j++ {
if s[j] == '.' {
if e > 0 || dot > 0 {
return false
}
dot++
} else if s[j] == 'e' || s[j] == 'E' {
if e > 0 || j == i || j == n-1 {
return false
}
e++
if s[j+1] == '+' || s[j+1] == '-' {
j++
if j == n-1 {
return false
}
}
} else if s[j] < '0' || s[j] > '9' {
return false
}
}
return true
}

• using System.Text.RegularExpressions;

public class Solution {
private readonly Regex _isNumber_Regex = new Regex(@"^\s*[+-]?(\d+(\.\d*)?|\.\d+)([Ee][+-]?\d+)?\s*\$");

public bool IsNumber(string s) {
return _isNumber_Regex.IsMatch(s);
}
}

• impl Solution {
pub fn is_number(s: String) -> bool {
let mut i = 0;
let n = s.len();

if let Some(c) = s.chars().nth(i) {
if c == '+' || c == '-' {
i += 1;
if i == n {
return false;
}
}
}
if let Some(x) = s.chars().nth(i) {
if
x == '.' &&
(i + 1 == n ||
(if let Some(m) = s.chars().nth(i + 1) { m == 'e' || m == 'E' } else { false }))
{
return false;
}
}

let mut dot = 0;
let mut e = 0;
let mut j = i;

while j < n {
if let Some(c) = s.chars().nth(j) {
if c == '.' {
if e > 0 || dot > 0 {
return false;
}
dot += 1;
} else if c == 'e' || c == 'E' {
if e > 0 || j == i || j == n - 1 {
return false;
}
e += 1;
if let Some(x) = s.chars().nth(j + 1) {
if x == '+' || x == '-' {
j += 1;
if j == n - 1 {
return false;
}
}
}
} else if !c.is_ascii_digit() {
return false;
}
}
j += 1;
}

true
}
}