Leetcode-Valid Number

LiBlog發表於2014-11-25
LeetCode

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.

Analysis:

The key point is to clarify which case is valid. I used a recursive method with several control switches. With such method, it is easy for us to definite different rules of feasibility.

Solution:

 1 public class Solution {
 2     public boolean isNumber(String s) {
 3         s = s.trim();
 4         //check whether there is 'e'.  
 5         int index = s.indexOf("e");
 6 
 7         if (index!=-1){
 8             String left = s.substring(0,index);
 9             String right = s.substring(index+1,s.length());
10             if (isNumberRecur(left,true,true,true,false)&&isNumberRecur(right,false,true,true,false))
11                 return true;
12             else return false;
13         } else
14             if (isNumberRecur(s,true,true,true,false)) return true;
15             else return false;
16         
17     }
18 
19     public boolean isNumberRecur(String s, boolean canBeDouble, boolean canHaveSymbol, boolean canZeroHead, boolean canBeNull){
20         if (!canBeNull && s.isEmpty()) return false;
21         if (canBeNull && s.isEmpty()) return true;
22         int index;
23         
24         //NOTE: check symbol before checking float!
25         //check positive symbol
26         index = s.indexOf("+");
27         if (index!=-1 && !canHaveSymbol) return false;
28         if (canHaveSymbol && index!=-1 && index!=0) return false;
29         if (canHaveSymbol && index==0)
30             return isNumberRecur(s.substring(1,s.length()),true,false,true,false);
31         
32         //check negative symbol
33         index = s.indexOf("-");
34         if (index!=-1 && !canHaveSymbol) return false;
35         if (canHaveSymbol && index!=-1 && index!=0) return false;
36         if (canHaveSymbol && index==0)
37             return isNumberRecur(s.substring(1,s.length()),true,false,true,false);
38 
39         index = s.indexOf(".");
40         if (canBeDouble && index!=-1){            
41             String left = s.substring(0,index);
42             String right = s.substring(index+1,s.length());
43             
44             //NOTE: this code is only for the case "3." to be true in leetcode while "." is invalid.
45             if (left.isEmpty() && right.isEmpty()) return false;
46             if (isNumberRecur(left,false,true,true,true) && isNumberRecur(right,false,false,true,true))
47                 return true;
48             else return false;
49         }
50 
51         if (!canBeDouble && index!=-1) return false;
52         
53         //check zero head.
54         if (!canZeroHead && s.charAt(0)=='0' && s.length()>1) return false;
55 
56         //check whether all chars are numbers.
57         for (int i=0;i<s.length();i++)
58             if (s.charAt(i)<'0' || s.charAt(i)>'9') return false;
59 
60         return true;
61     }
62             
63 }

 

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