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## Problem

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

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

A 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.

## Code

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) == '.') {
// 小数点出现在e的后边 || 之前出现了小数点
// "e" false,  ".e1" false, "3.e" false, "3.e1" true
// "+e1" false, "1+e" false
if(eSeen || pointSeen) {
return false;
}

pointSeen = true;
} else if(s.charAt(i) == 'e' || s.charAt(i) == 'E') {
// 已经出现了e
// 还没有出现数字
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;
}
}