Valid Number Java

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.

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.

作为一道目前AP绿最低的一道题，着实蛋疼，感觉就是需要考虑的情况比较多，自己尝试多次之后毅然去了讨论区，找了一份如下代码：

源码地址：https://leetcode.com/discuss/26682/clear-java-solution-with-ifs

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('0' <= s.charAt(i) && 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;
}

并且附上原文解释：

We start with trimming.

If we see [0-9] we reset the number flags.

We can only see . if we didn’t see e or ..

We can only see e if we didn’t see e but we did see a number. We reset numberAfterE flag.

We can only see + and - in the beginning and after an e

any other character break the validation.

At the and it is only valid if there was at least 1 number and if we did see an e then a number after it as well.

So basically the number should match this regular expression:

[-+]?[0-9]*(.[0-9]+)?(e[-+]?[0-9]+)?