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.
1 public class Solution { 2 public boolean isNumber(String s) { 3 boolean result = true; 4 s = s.trim();//去掉前面和后面的空格 5 if(0 != s.length() && (s.charAt(0) == '+' || s.charAt(0) == '-')) 6 s = s.substring(1, s.length()); 7 if(0 == s.length()) 8 return false; 9 boolean flag_point = false; 10 boolean flag_e = false;//记录小数点和字母e 11 12 13 for(int i = 0; i < s.length(); i++){ 14 char ch = s.charAt(i);//第i个字符 15 if('.' == ch){ 16 if(flag_e || flag_point || !((i + 1 < s.length() && s.charAt(i + 1) >= '0' && s.charAt(i + 1) <= '9')|| 17 (i - 1 >= 0 && s.charAt(i - 1) >= '0' && s.charAt(i - 1) <= '9'))//e出现过就不能出现小数点了 18 ) 19 { 20 return false; 21 } 22 flag_point = true; 23 }//小数点后面或者前面有数字即可 24 if('e' == ch){ 25 if(flag_e || 26 !((i + 1 < s.length() && (s.charAt(i + 1) >= '0' && s.charAt(i + 1) <= '9' || s.charAt(i + 1) == '+'|| s.charAt(i + 1) == '-'))&& 27 (i - 1 >= 0 && (s.charAt(i - 1) >= '0' && s.charAt(i - 1) <= '9' || s.charAt(i - 1) == '.')))) 28 { 29 return false; 30 } 31 flag_e = true; 32 }//小数点和e的个数 33 if(flag_e && (ch == '+' || ch == '-')){ 34 if(!(s.charAt(i - 1) == 'e') || 35 !(i + 1 < s.length() && (s.charAt(i + 1)>= '0' && s.charAt(i + 1) <= '9'))) 36 return false; 37 }//+ 出现过e 38 else if(!((ch >= '0' && ch <= '9') || ch == '.' || ch == 'e')){ 39 return false; 40 } 41 } 42 43 return result; 44 } 45 }
还是要先看一下哪一些是合理的，可以先百度一下。在实现，要不然后面要针对测试用例一个个修改，严重拉低通过率