Leetcode: valid number solution report

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 shoshould gather all requirements up front before implementing one.

Solution 1:

We set three flags:

1. Num

2. Exp

3. Dot

There are the following situations:

(1). If e appears, there must be digit first, and there must be no E. There must be digit later.

(2). If it appears, it is a decimal number, so there cannot be. And E

(3). If + appears,-it must be the first, or the previous one is E, for example, "005047e + 6"

The actual code implementation is as follows, which is quite concise.

Thanks to the answer provider: http://www.ninechapter.com/solutions/valid-number/

Wow!

 1 public class Solution { 2     public boolean isNumber(String s) { 3         if (s == null) { 4             return false; 5         } 6          7         // cut the leading spaces and tail spaces. 8         String sCut = s.trim(); 9         10         /*11         Some examples:12         "0" => true 13         " 0.1 " => true14         "abc" => false15         "1 a" => false16         "2e10" => true17         */18         19         int len = sCut.length();20         21         boolean num = false;22         boolean exp = false;23         boolean dot = false;24         25         for (int i = 0; i < len; i++) {26             char c = sCut.charAt(i);27             if (c == ‘e‘) {28                 if (!num || exp) {29                     return false;30                 }31                 exp = true;32                 num = false; // Should be: 2e2 , so there should be number follow "e"33             } else if (c <= ‘9‘ && c >= ‘0‘) {34                 num = true;35             } else if (c == ‘.‘) {36                 if (exp || dot) { // can‘t be: e0.2 can‘t be: ..37                     return false;38                 }39                 dot = true;40             } else if (c == ‘+‘ || c == ‘-‘) {41                 if (i != 0 && sCut.charAt(i - 1) != ‘e‘) { // filter : " 005047e+6", this is true.42                     return false;43                 }44             } else {45                 // invalid character.46                 return false;47             }48         }49         50         return num;51     }52 }

View code

Please go to the home page of Jun GitHub: https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/string/IsNumber.java

 

You can also use regular expressions:

Recommended Tutorials:

Http://developer.51cto.com/art/200912/166310.htm

Http://luolei.org/2013/09/regula-expression-simple-tutorial/

Http://net.tutsplus.com/tutorials/php/regular-expressions-for-dummies-screencast-series/

Http://deerchao.net/tutorials/regex/regex.htm

The homepage is not written, because it is too complicated to write regular expressions during the interview.

Answers to the regular expressions written by a great God:

Http://blog.csdn.net/fightforyourdream/article/details/12900751? Reload

Leetcode: valid number solution report