# 65.Valid-Number

## 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

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

class Solution {
public boolean isNumber(String s) {
return s.trim().matches("[-+]?(\\d+\\.?|\\.\\d+)\\d*(e[-+]?\\d+)?");
}
}
Last modified 3yr ago