*Valid Number

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.

解法一：

/**
* 本代码由九章算法编辑提供。没有版权欢迎转发。
* - 九章算法致力于帮助更多中国人找到好的工作，教师团队均来自硅谷和国内的一线大公司在职工程师。
* - 现有的面试培训课程包括：九章算法班，系统设计班，BAT国内班
* - 更多详情请见官方网站：http://www.jiuzhang.com/
*/

// Non-regex version

public class Solution {
public boolean isNumber(String s) {
int len = s.length();
int i = 0, e = len - 1;
while (i <= e && Character.isWhitespace(s.charAt(i))) i++;
if (i > len - 1) return false;
while (e >= i && Character.isWhitespace(s.charAt(e))) e--;
// skip leading +/-
if (s.charAt(i) == '+' || s.charAt(i) == '-') i++;
boolean num = false; // is a digit
boolean dot = false; // is a '.'
boolean exp = false; // is a 'e'
while (i <= e) {
char c = s.charAt(i);
if (Character.isDigit(c)) {
num = true;
}
else if (c == '.') {
if(exp || dot) return false;
dot = true;
}
else if (c == 'e') {
if(exp || num == false) return false;
exp = true;
num = false;
}
else if (c == '+' || c == '-') {
if (s.charAt(i - 1) != 'e') return false;
}
else {
return false;
}
i++;
}
return num;
}
}

解法二：

CleanCode Version: better logic

public class Solution {
public boolean isNumber(String s)
{
int i=0;
int n=s.length();
while(i<n&&Character.isWhitespace(s.charAt(i)))i++;
if(i<n&&(s.charAt(i)=='+'||s.charAt(i)=='-'))i++;
boolean isNumeric = false;
while(i<n&&Character.isDigit(s.charAt(i)))
{
i++;
isNumeric = true;
}
if(i<n&&s.charAt(i)=='.')
{
i++;
while(i<n&&Character.isDigit(s.charAt(i)))
{
i++;
isNumeric =  true;
}
}
if(isNumeric&&i<n&&s.charAt(i)=='e')
{
i++;
isNumeric = false;
if(i<n&&(s.charAt(i)=='+'||s.charAt(i)=='-'))i++;
while(i<n&&Character.isDigit(s.charAt(i)))
{
i++;
isNumeric =  true;
}
}
while(i<n&&Character.isWhitespace(s.charAt(i)))i++;
return i==n&&isNumeric;
}
}