> For the complete documentation index, see [llms.txt](https://wentao-shao.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://wentao-shao.gitbook.io/leetcode/numbers/65.valid-number.md).

# 65.Valid-Number

## 65. Valid Number

## 题目地址

<https://leetcode.com/problems/swap-nodes-in-pairs/>

## 题目描述

```
Validate if a given string can be interpreted as a decimal number.

Some examples:
"0" => true
" 0.1 " => true
"abc" => false
"1 a" => false
"2e10" => true
" -90e3   " => true
" 1e" => false
"e3" => false
" 6e-1" => true
" 99e2.5 " => false
"53.5e93" => true
" --6 " => false
"-+3" => false
"95a54e53" => false

Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. However, here is a list of characters that can be in a valid decimal number:

Numbers 0-9
Exponent - "e"
Positive/negative sign - "+"/"-"
Decimal point - "."
Of course, the context of these characters also matters in the input.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.
```

## 代码

### Approach #1

```java
class Solution {
  public boolean isNumber(String s) {
        s = s.trim()
    boolean pointSeen = false;
    boolean eSeen = false;
    boolean numberSeen = false;
    boolean numberAfterE = true;

    for (int i = 0; i < s.length(); i++) {
      if (s.charAt(i) >= '0' && s.charAt(i) <= '9') {
        numberSeen = true;
        numberAfterE = true;
      } else if (s.charAt(i) == '.') {
        if (eSeen || pointSeen) {
          return false;
        }
        pointSeen = true;
      } else if (s.charAt(i) == 'e') {
        if (eSeen || !numberSeen) {
          return false;
        }
        numberAfterE = false;
        eSeen = true;
      } else if (s.charAt(i) == '-' || s.charAt(i) == '+') {
        if (i != 0 && s.charAt(i - 1) != e) {
          return false;
        } 
      } else {
        return false;
      }
    }

    return numberSeen && numberAfterE;
  }
}
```

### Approach #2

```java
class Solution {
  public boolean isNumber(String s) {
    return s.trim().matches("[-+]?(\\d+\\.?|\\.\\d+)\\d*(e[-+]?\\d+)?");
  }
}
```