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.
这道题就是将所有的情况都想出来，然后按条件写好即可
我们首先trim掉头尾的空格，和开头的+和-
然后剩下的所有字符里只能有. e +/-最多各一个和其他的数字
然后就是每种情况分别写好即可。注意e不可以开头结尾，.不可以结尾，还有其他的一些，他们之间的位置关系，列清楚即可通过
public class Solution {
public boolean isNumber(String s) {
s = s.trim(); // remove trailing and heading spaces
if (s.length() == 0)
return false;
StringBuilder in = new StringBuilder(s);
if ((s.charAt(0)) == '+' || (s.charAt(0) == '-')) // remove the head +/-
in.deleteCharAt(0);
int dot = -1;
int e = -1;
int sign = -1; // decimal point and e position and sign position
for (int i = 0; i < in.length(); i++)
{
if (in.charAt(i) == '.' && dot == -1)
dot = i;
else if (in.charAt(i) == 'e' && e == -1)
e = i;
else if (in.charAt(i) =='-' || in.charAt(i) == '+')
sign = i;
else if (!Character.isDigit(in.charAt(i)))
return false;
}
if (dot == -1 && e == -1 && sign == -1) //xxx
{
return true;
}
else if (dot != -1 && e == -1 && sign == -1 ) //xxx.xxx
{
if (dot == 0 && in.length() == 1) //xxx. is legal, but . itself is not
return false;
return true;
}
else if (dot == -1 && e != -1 && sign == -1) //xxexx
{
if ( e == in.length() - 1 || e == 0) //xxe is not legal
return false;
return true;
}
else if (dot != -1 && e != -1 && sign == -1)
{
if ((dot == 0 && dot == e-1) || dot == in.length() - 1 || e == in.length() - 1 || e == 0 || dot > e) // xx.e is not legal .exxx is not legal
return false;
return true;
}
else if (dot != -1 && e != -1 && sign != -1)
{
if (dot == in.length() - 1 || e == in.length() -1 || sign != e+1 || sign == in.length() - 1 || e == 0)
return false;
return true;
}
else if (dot == -1 && e != -1 && sign != -1)
{
if (sign != e+1 || sign == in.length() - 1 || e == 0)
return false;
return true;
}
else
return false;
}
}
Second Submission:
Use regular expression, but be careful with the matching
public class Solution {
public boolean isNumber(String s) {
String regex = "[-+]?((\\d+\\.?)|(\\d*\\.?\\d+))(e[+-]?\\d+)?";
return s.trim().matches(regex);
}
}
public class Solution {
public boolean isNumber(String s) {
s = s.trim(); // remove the heading and trailing space
if (s.length() == 0)
return false;
if (s.charAt(0) == '-' || s.charAt(0) == '+')
s = s.substring(1);
if (s.length() == 0)
return false;
// find the position of . and e
int dot = -1;
int e = -1;
int sign = -1; //
for (int i = 0; i < s.length(); i++) {
if (!Character.isDigit(s.charAt(i))) {
if (s.charAt(i) == '.') {
if (dot == -1)
dot = i;
else
return false; // can not be more than one .
continue;
}
if (s.charAt(i) == 'e') {
if (e == -1)
e = i;
else
return false; // can not be more than one e
continue;
}
if (s.charAt(i) == '+' || s.charAt(i) == '-') {//xx.xxe+/-xx
if (e != -1 && i == e+1) {
continue;
}
}
return false;
}
}
if (dot == -1 && e == -1) //XXX all numbers
return true;
if (dot != -1 && e == -1) {//XXX.XXX
if (s.length() == 1)
return false;
}
if (dot == -1 && e != -1) { //XXXeXXX
if (e == 0 || e == s.length()-1)
return false;
if (s.charAt(e+1) == '-' || s.charAt(e+1) == '+') { //xxe+xx
if (e+1 == s.length() - 1)
return false;
}
}
if (dot != -1 && e != -1) {//XXX.XXeXX
if (dot > e || e == s.length() - 1) // XXXeX. XX.Xe
return false;
if (dot == 0 && dot == e-1)
return false;
if (s.charAt(e+1) == '-' || s.charAt(e+1) == '+') { //x.xe+xx
if (e+1 == s.length() - 1)
return false;
}
}
return true;
}
}
A cleaner answer: skip the possible spaces, Signs, digits and . and e in proper order.
public class Solution {
public boolean isNumber(String s) {
if (s.length() == 0)
return false;
int start = 0;
start += skipSpaces(s, start); //skip space if exist
if (start == s.length()) //the string has only " " characters
return false;
start += skipSign(s, start);
int d1 = skipDigits(s, start);
start += d1;
if (start == s.length()) // xxx only
return true;
if (s.charAt(start) == '.') //xxx.xx
start++;
int d2 = skipDigits(s, start);
start += d2;
if (d1 == 0 && d2 == 0)
return false;
if (start == s.length()) // xxx.xx
return true;
if (s.charAt(start) == 'e' || s.charAt(start) == 'E') {//xxxexxx
start++;
start += skipSign(s, start);
int d3 = skipDigits(s, start);
if (d3 == 0)
return false;
start += d3;
}
start += skipSpaces(s, start);
return (start == s.length());
}
public int skipSign(String s, int start) {
if (start < s.length() && (s.charAt(start) == '+' || s.charAt(start) == '-'))
return 1;
return 0;
}
public int skipDigits(String s, int start) {
int len = 0;
while (start < s.length() && Character.isDigit(s.charAt(start))) {
len++;
start++;
}
return len;
}
public int skipSpaces(String s, int start) {
//increment until space was not found
int len = 0;
while(start < s.length() && (s.charAt(start) == ' ')) {
len++;
start++;
}
return len;
}
}
Hello World 