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.

Solution:

  1. public class Solution {
  2.   public boolean isNumber(String s) {
  3.     if (s == null) return false;
  5.     s = s.trim();
  6.     int n = s.length();
  8.     if (n == 0) return false;
  10.     // Define flags
  11.     int signCount = 0;
  13.     boolean hasE = false;
  14.     boolean hasNum = false;
  15.     boolean hasPoint = false;
  17.     // Go through the characters
  18.     for (int i = 0; i < n; i++) {
  19.       char c = s.charAt(i);
  21.       // invalid character
  22.       if (!isValid(c)) return false;
  24.       // digit
  25.       if (c >= '0' && c <= '9') hasNum = true;
  27.       // e or E
  28.       if (c == 'e' || c == 'E') {
  29.         // e 之前一定要有数字
  30.         if (hasE || !hasNum) return false;
  31.         // e 不能作为最后一个
  32.         if (i == n - 1) return false;
  34.         hasE = true;
  35.       }
  37.       // decimal place
  38.       if (c == '.') {
  39.         // 小数点不能出现在 e 之后
  40.         if (hasPoint || hasE) return false;
  41.         // 小数点如果在最后一位出现，那么前面必须有要数字
  42.         if (i == n - 1 && !hasNum) return false;
  44.         hasPoint = true;
  45.       }
  47.       // signs
  48.       if (c == '+' || c == '-') {
  49.         // 不允许超过两个符号出现
  50.         if (signCount == 2) return false;
  51.         // 不允许符号在最后出现
  52.         if (i == n - 1) return false;
  53.         // 符号出现在中间的前提是前面有 e
  54.         if (i > 0 && !hasE) return false;
  56.         signCount++;
  57.       }
  58.     }
  60.     return true;
  61.   }
  63.   boolean isValid(char c) {
  64.     return c == '.' || c == '+' || c == '-' || c == 'e' || c == 'E' || c >= '0' && c <= '9';
  65.   }
  66. }