Question

65. Valid Number

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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:
  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

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;
  }
};

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
}

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
  }
}

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);
  }
}