[Leetcode] valid number

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.

Algorithm ideas:

Exclusion: If all illegal conditions occur, false is returned, and true is returned after scanning.

Illegal:

Symbol: +,-can appear only twice at most, and the second time must appear after E/E. In addition, there must be no number before the first plus sign (+/-), and the second must be followed by a number after E;

E: it can only appear once at most. Besides, numbers must be placed before and after each vertex;

.: It can only appear once at most, and cannot appear after E. If it is the last digit of a string, there must be a number before it;

[Post-Submission error supplement] at the first submission, "E" reports an error. The boundary is not considered.

In short, this question is not difficult, but tedious:

I hope you can learn from the following error cases: (a lot of cases)

005047e + 6 true

3. True

. 1 true

-. False

6 + 1 false

. 1 false

0e false

The Code is as follows:

 1 public class Solution { 2     public boolean isNumber(String s) { 3         if(s == null || s.trim().length() == 0) return false; 4         s = s.trim(); 5         char[] charArray = {‘0‘,‘1‘,‘2‘,‘3‘,‘4‘,‘5‘,‘6‘,‘7‘,‘8‘,‘9‘,‘.‘,‘e‘,‘E‘,‘+‘,‘-‘}; 6         Set<Character> set = new HashSet<Character>(); 7         for(char c : charArray){ 8             set.add(c); 9         }10         boolean hasE = false, hasPoint = false, hasNum = false;11         int operaterCount = 0;12         for(int i = 0; i < s.length(); i++){13             if(!set.contains(s.charAt(i))) return false;14             if(s.charAt(i) <= ‘9‘ && s.charAt(i) >= ‘0‘) hasNum = true;15             if(s.charAt(i) == ‘E‘ || s.charAt(i) == ‘e‘){16                 if(i == 0 || i == s.length() - 1) return false;17                 if(!hasE && hasNum) hasE = true;18                 else return false;19             }20             if(s.charAt(i) == ‘.‘){21                 if(i == 0 && 1 == s.length()) return false;22                 if(i == s.length() - 1 && !hasNum) return false;23                 if(!hasPoint && !hasE) hasPoint = true;24                 else return false;25             }26             if(s.charAt(i) == ‘-‘ || s.charAt(i) == ‘+‘){27                 if(i == s.length() - 1 || (i != 0 && s.charAt(i - 1) != ‘e‘ && s.charAt(i - 1) != ‘E‘)) return false;28                 if(operaterCount == 2) return false;29                 if(operaterCount == 1 && !hasE && !hasNum) return false;30                 operaterCount++;31             }32         }33         return true;34     }35 }