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Posts tagged as “digit”

花花酱 LeetCode 65. Valid Number

valid number can be split up into these components (in order):

  1. decimal number or an integer.
  2. (Optional) An 'e' or 'E', followed by an integer.

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

Example 4:

Input: s = ".1"
Output: true

Constraints:

  • 1 <= s.length <= 20
  • s consists of only English letters (both uppercase and lowercase), digits (0-9), plus '+', minus '-', or dot '.'.

Solution: Rule checking

Time complexity: O(n)
Space complexity: O(1)

C++

花花酱 LeetCode 728. Self Dividing Numbers

Problem:

self-dividing number is a number that is divisible by every digit it contains.

For example, 128 is a self-dividing number because 128 % 1 == 0128 % 2 == 0, and 128 % 8 == 0.

Also, a self-dividing number is not allowed to contain the digit zero.

Given a lower and upper number bound, output a list of every possible self dividing number, including the bounds if possible.

Example 1:

Note:

  • The boundaries of each input argument are 1 <= left <= right <= 10000.




Idea:

Brute Force

Time Complexity: O(n)

Space Complexity: O(1)

 

Solution:

C++

String

 

Related Problems: