最复杂的情况:-5.23E-23(只是一个例子)
以此为例子可以分为:碰到“+-”号,碰到数字0-9,碰到“.”,碰到“E”或“e”来分析
即A.BEC(这里的E还可以为e),A,C都可以是负数,B是无符号数。所以逐个封装(运行通不过,佛了)
public class Solution {
private int index = 0;
public boolean isNumeric(char[] str) {
if (str.length < 1)
return false;
boolean flag = scanInteger(str);
if (index < str.length && str[index] == '.') {
index++;
flag = scanUnsignedInteger(str) || flag;
}
if (index < str.length && (str[index] == 'E' || str[index] == 'e')) {
index++;
flag = flag && scanInteger(str);
}
return flag && index == str.length;
}
private boolean scanInteger(char[] str) {
if (index < str.length && (str[index] == '+' || str[index] == '-') )
index++;
return scanUnsignedInteger(str);
}
private boolean scanUnsignedInteger(char[] str) {
int start = index;
while (index < str.length && str[index] >= '0' && str[index] <= '9')
index++;
return start < index; //是否存在整数
}
}
正则表达式:
java.lang.String
java.util.regex.Pattern
java.util.regex.Matcher
public class Solution {
public boolean isNumeric(char[] str) {
String s=String.valueOf(str);
//return s.matches("[+-]?[0-9]*(.[0-9]*)?([eE][+-]?[0-9]+)?");
return s.matches("[+-]?[0-9]*(\\.[0-9]*)?([eE][+-]?[0-9]+)?");
}
}