Difficulty: Hard
More:【目录】LeetCode Java实现
Description
Validate if a given string can be interpreted as a decimal number.
Some examples:"0" => true" 0.1 " => true"abc" => false"1 a" => false"2e10" => true" -90e3 " => true" 1e" => false"e3" => false" 6e-1" => true" 99e2.5 " => false"53.5e93" => true" --6 " => false"-+3" => false"95a54e53" => false
Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. However, here is a list of characters that can be in a valid decimal number:
- Numbers 0-9
- Exponent - "e" or "E"
- Positive/negative sign - "+"/"-"
- Decimal point - "."
Of course, the context of these characters also matters in the input.
Intuition
Method1: A valid number is in the form of A.B e/E A (A: integer, B: unsigned integer), so it is helpful to break the problem down to several components that can be solved individually. Detailed solution refer to: 表示数值的字符串
Method2: Use some flags(eSeen, pointSeen, isNum) while scan each character in the String. The solution is shown below.
Solution
public boolean isNumber(String s) {
if(s==null || s.length()<=0)
return false;
s=s.trim();
boolean isNum=false;
boolean pointSeen=false;
boolean eSeen=false;
for(int i=0;i<s.length();i++){
if(s.charAt(i)=='+'||s.charAt(i)=='-'){
if(i!=0 && s.charAt(i-1)!='e' && s.charAt(i-1)!='E')
return false;
}else if(Character.isDigit(s.charAt(i))){
isNum=true;
}else if(s.charAt(i)=='.'){
if(eSeen || pointSeen)
return false;
pointSeen=true;
}else if(s.charAt(i)=='e' || s.charAt(i)=='E' ){
if(eSeen || !isNum)
return false;
eSeen=true;
isNum=false;
}else
return false;
}
return isNum;
}
Complexity
Time complexity : O(n)
Space complexity : O(1)
What I've learned
1. Ought to make the best of flags. Learn to use flags well.
More:【目录】LeetCode Java实现