# 65_Valid_Number
###### tags: `leetcode`
## Problem Statement
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:
At least one digit, followed by a dot '.'.
At least one digit, followed by a dot '.', followed by at least one digit.
A dot '.', followed by at least one digit.
An integer can be split up into these components (in order):

(Optional) A sign character (either '+' or '-').
At least one digit.
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
- Give out each character in the string the constraints that they need to follow.
- Each element can be seperated into 2 main parts.
- ```e``` or ```E``` have already appeared.
- Not yet.
- If appeared, Only ```integer``` is allowed to be a valid one, else can be a float.
- The 1^st^ case: ```+/-```
- ```cnt``` is 1 or 4 depends on ```e```.
- If it is the last one, not the first and the last one is not ```e```, only '.': false.

if (s[i]== '+'|| s[i]== '-')
{
    if (i== s.length()- 1|| (i!= 0&& cnt!= 3)|| s.length()== 1)
        return 0;
    if (cnt== 3)
        cnt= 4;
    else
        cnt= 1;
}

- The 2^nd^ case: ```0~9```
- ```cnt``` is 2 or 5 depends on ```e```.

else if (s[i]- '0'> -1&& s[i]- '0'< 10)
{
    if (cnt> 2)
        cnt= 5;
    else
        cnt= 2;
}

- The 3^rd^ case: ```e/E```
- ```cnt``` is 3.
- If the last one is not a number, it is the last one or the first: false.

else if (s[i]== 'e'|| s[i]== 'E')
{
    if (cnt!= 2|| i== s.length()- 1|| i== 0)
        return 0;
    cnt= 3;
}

- The last case: ```.```
- If there has already one dot, only ```.```, ```e/ E``` has appeared, It is the last one and hte last one is not a number: false.

else if (s[i]== '.')
{
    if (dot|| s.length()== 1|| cnt> 2|| (i== s.length()- 1&& cnt% 3!= 2))
        return 0;
    dot= 1;
}