65. Valid Number

Description

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:
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):

(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 '.'.

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.

Python Code

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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

Java Code

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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;
    }
}

C++ Code

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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;
    }
};

Go Code

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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
}

Rust Code

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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
    }
}

C# Code

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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);
    }
}

65. Valid Number#

Description#

Solutions#

Solution 1: Case Discussion#

65. Valid Number

Description

Solutions

Solution 1: Case Discussion