> For the complete documentation index, see [llms.txt](https://hao-fu-1.gitbook.io/oj/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://hao-fu-1.gitbook.io/oj/string/valid-number.md).

# Valid Number

https\://leetcode.com/problems/valid-number/description/

> Validate if a given string is numeric.
>
> Some examples:
>
> "0" => true
>
> " 0.1 " => true
>
> "abc" => false
>
> "1 a" => false
>
> "2e10" => true
>
> Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one.

## Thoughts

什么样是合法的数:\
1\. 首尾可以出现无数个空格\
2\. 第一个字符可以是"+/-"\
3\. 之后必须是数字或者".".\
4\. 之后可以是'e', 如果是'e'后面可以接"+/-"\
5\. 之后必须是数字.\
像.3, -3., -3.e-2啥的都是合法的, 但'.'和'e2'之类的就不合法.\
根据这些顺序的规则, 我们用hasDigit判断是否有数字, 如果遇到e就置为false, 后面再有数字置为true。最后判断是否i到了N且hasDigit。

## Code

```
/*
 * @lc app=leetcode id=65 lang=cpp
 *
 * [65] Valid Number
 */
class Solution {
public:
    bool isNumber(string s) {
        const int N = s.size();
        int i = 0;
        // trim
        while (i < N && s[i] == ' ') ++i;
        // the sign
        if (i < N && (s[i] == '+' || s[i] == '-')) ++i;
        // digits until
        bool digit = false;
        while (i < N && (s[i] >= '0' && s[i] <= '9')) {
            digit = true;
            ++i;
        }
        // the dot
        if (i < N && s[i] == '.') ++i;
        // continue
        while (i < N && (s[i] >= '0' && s[i] <= '9')) {
            digit = true;
            ++i;
        }
        // 'e'
        if (i < N && s[i] == 'e' && digit) {
            digit = false;
            ++i;
            if (i < N && (s[i] == '+' || s[i] == '-')) ++i;
        }
        // continue
        while (i < N && (s[i] >= '0' && s[i] <= '9')) {
            digit = true;
            ++i;
        }
        // trim
        while (i < N && s[i] == ' ') ++i;
        return (i == N) && digit;
    }
};


```

## Analysis

时间复杂度O(N).